Exercices : Identités remarquable – 3ac

$\begin{aligned} \mathbf{1.}\quad A &=2(x+5) =2 x+10\\ \mathbf{2.}\quad B &= \left( {5- x} \right)\left( {7 + x} \right)= 35 + 5x- 7x- {x^2}={- {x^2}- 2x + 35}\\ \mathbf{3.}\quad C &= \frac{2}{3}\left( {5 + 7x} \right)- \frac{1}{2}\left( {- x + 1} \right)\\ &= \frac{{10}}{3} + \frac{{14}}{3}x + \frac{1}{2}x- \frac{1}{2}\\ &= \left( {\frac{{14}}{3} + \frac{1}{2}} \right)x + \left( {\frac{{10}}{3}- \frac{1}{2}} \right)\\ &= \left( {\frac{{28}}{6} + \frac{3}{6}} \right)x + \left( {\frac{{20}}{6}- \frac{3}{6}} \right)\\ &={ \frac{{31}}{6}x + \frac{{17}}{6}}\\ \mathbf{3.}\quad D &= \left( {x + 3} \right)\left( {{x^2}- 3x + 5} \right)\\ &= {x^3}- 3{x^2} + 5x + 3{x^2}- 9x + 15\\ &={ {x^3}- 4x + 15} \end{aligned}$

$\begin{aligned} \mathbf{1.}\quad A &= {\left( {x + 3} \right)^2}= {x^2} + 2 \times x \times 3 + {3^2}={ {x^2} + 6x + 9}\\ \mathbf{2.}\quad B &= {\left( {3x-1} \right)^2}= {\left( {3x} \right)^2}-2 \times 3x \times 1 + {1^2}={ 9{x^2}-6x + 1}\\ \mathbf{3.}\quad C &= 5{\left( {1-x} \right)^2}= 5\left( {{1^2}-2 \times 1 \times x + {x^2}} \right)= 5\left( {1-2x + {x^2}} \right)={ 5-10x + 5{x^2}}\\ \mathbf{4.}\quad D &= \left( {3x + 7} \right)\left( {3x-7} \right) + 4{\left( {x-\frac{1}{2}} \right)^2}\\ &= {\left( {3x} \right)^2}-{7^2} + 4\left( {{x^2}-2 \times x \times \frac{1}{2} + \left( \frac{1}{2} \right)^2} \right)\\ &= 9{x^2}-49 + 4\left( {{x^2}-x + \frac{1}{4}} \right)\\ &= 9{x^2}-49 + 4{x^2}-4x + 1\\ &={ 13{x^2}-4x-48} \end{aligned}$

$\begin{aligned} \mathbf{1.}\quad A &= ab + 5b={ b\left( {a + 5} \right)}\\ \mathbf{2.}\quad B &= 12x + 18={ 6\left( {2x + 3} \right)}\\ \mathbf{3.}\quad C &= 5x-{x^2}={ x\left( {5-x} \right)}\\ \mathbf{4.}\quad D &= x + 5{x^2} + 11{x^3}={ x\left( {1 + 5x + 11{x^2}} \right)}\\ \mathbf{5.}\quad E &= 5\left( {x + 1} \right) + {\left( {x + 1} \right)^2}\\ &= \left( {x + 1} \right)\left( {5 + \left( {x + 1} \right)} \right)\\ &= \left( {x + 1} \right)\left( {5 + x + 1} \right)\\ &={ \left( {x + 1} \right)\left( {x + 6} \right)}\\ \mathbf{6.}\quad F &= \left( {x-3} \right)\left( {x + 7} \right)-\left( {5-x} \right)\left( {x-3} \right)\\ &= \left( {x-3} \right)\left[ {\left( {x + 7} \right)-\left( {5-x} \right)} \right]\\ &= \left( {x-3} \right)\left( {x + 7-5 + x} \right)\\ &={ \left( {x-3} \right)\left( {2x+2} \right)}={ 2\left( {x-3} \right)\left( {x+1} \right)} \end{aligned}$

$\begin{aligned} \mathbf{1.}\quad A &= {x^2}-49 + x\left( {x-7} \right)\\ &= \left( {{x^2}-{7^2}} \right) + x\left( {x-7} \right)\\ &= \left( {x-7} \right)\left( {x + 7} \right) + x\left( {x-7} \right)\\ &= \left( {x-7} \right)\left[ {\left( {x + 7} \right) + x} \right]\\ &={ \left( {x-7} \right)\left( {2x + 7} \right)} \end{aligned}$

$\begin{aligned} \mathbf{2.}\quad B &= {x^2} + 4x + 4\\ &= {x^2} + 2 \times x \times 2 + {2^2}\\ &={ {\left( {x + 2} \right)^2}} \end{aligned}$

$\begin{aligned} \mathbf{3.}\quad C &= {x^2}-\frac{9}{{121}}\\ &= {x^2}-{\left( {\frac{3}{{11}}} \right)^2}\\ &={ \left( {x-\frac{3}{{11}}} \right)\left( {x + \frac{3}{{11}}} \right)} \end{aligned}$

$\begin{aligned} \mathbf{4.}\quad D &= {\left( {2x-3} \right)^2}-{\left( {x + 1} \right)^2}\\ &= \left[ {\left( {2x-3} \right)-\left( {x + 1} \right)} \right]\left[ {\left( {2x-3} \right) + \left( {x + 1} \right)} \right]\\ &= \left( {2x-3-x-1} \right)\left( {2x-3 + x + 1} \right)\\ &= {\left( {x-4} \right)\left( {3x-2} \right)} \end{aligned}$

$\begin{aligned} \mathbf{5.}\quad E &= \frac{{{x^2}}}{8}-8\\ &= \frac{1}{8}\left( {{x^2}-{8^2}} \right)\\ &= {\frac{1}{8}\left( {x-8} \right)\left( {x + 8} \right)} \end{aligned}$

$\begin{aligned} \mathbf{6.}\quad F &=-7{x^2} + 14x-7\\ &=-7\left( {{x^2}-2x + 1} \right)\\ &= {-7{\left( {x-1} \right)^2}} \end{aligned}$

$\begin{aligned} \mathbf{7.}\quad G &= {x^{12}}-1 + 5\left( {{x^6}-1} \right)\\ &= {\left( {{x^6}} \right)^2}-1^2 + 5\left( {{x^6}-1} \right)\\ &= \left( {{x^6}-1} \right)\left( {{x^6} + 1} \right) + 5\left( {{x^6}-1} \right)\\ &= \left( {{x^6}-1} \right)\left( {{x^6} + 1 + 5} \right)\\ &= {\left( {{x^6}-1} \right)\left( {{x^6} + 6} \right)} \end{aligned}$

$\begin{aligned} \mathbf{8.}\quad H &= {x^2}-x-6\\ &= \left( {{x^2}-4} \right)-\left( {x + 2} \right)\\ &= \left( {x-2} \right)\left( {x + 2} \right)-\left( {x + 2} \right)\\ &= \left( {x + 2} \right)\left( {x-2-1} \right)\\ &= {\left( {x + 2} \right)\left( {x-3} \right)} \end{aligned}$

1. Développons et réduisons $P$.
   $$\begin{aligned} P &= {(x-1)^2}-(x-1)(3x-4)\\ &= \left[ {{x^2}-2 \times x \times 1 + {1^2}} \right]-\left[ {3{x^2}-4x-3x + 4} \right]\\ &= {x^2}-2x + 1-3{x^2} + 4x + 3x-4\\ &={-2{x^2} + 5x-3} \end{aligned}$$
2. Factoriser $P$.
   $$\begin{aligned} P &= {(x-1)^2}-(x-1)(3x-4)\\ &= \left( {x-1} \right)\left[ {\left( {x-1} \right)-\left( {3x-4} \right)} \right]\\ &= \left( {x-1} \right)\left[ {x-1-3x + 4} \right]\\ &={ \left( {x-1} \right)\left( {-2x + 3} \right)} \end{aligned}$$
3. Calculons $P$ pour $x=1$ puis pour $x=\dfrac{1}{3}$.Pour $x=1$, on choisit l’expression $P= \left( {x-1} \right)\left( {-2x + 3} \right)$, on a : $$P= \left( {1-1} \right)\left( {-2\times 1 + 3}\right)=0 \times 1 ={0}$$ Pour $x=\dfrac{1}{3}$, on choisit l’expression $P=-2{x^2} + 5x-3$, on a : $$P=-2{\left(\dfrac{1}{3}\right)^2}+5\times \dfrac{1}{3}-3 = \dfrac{-2}{9}+\dfrac{5}{3}-3 =\dfrac{-2}{9}+\dfrac{15}{9}-\dfrac{27}{9}={\dfrac{-14}{9}}$$
4. On a $P=0$, alors $(x-1)^2-(x-1)(3x-4)=0$, donc $\left( {x-1} \right)\left( {-2x + 3} \right)=0$, alors $x-1=0$ ou $-2x+3=0$, alors $x=1$ ou $x=\dfrac{3}{2}$.D’où l’équation admet deux solutions sont : $1$ et $\dfrac{3}{2}$.

1. On a :
   $$\begin{aligned} B&= (2x-5)^2-36\\ &= (2x-5)^2-6^2\\ &= (2x-5-6)(2x-5+6)\\ &= (2x-11)(2x+1). \end{aligned}$$
2. On a :
   $$\begin{aligned} B -2A &= {\left( {2x -5} \right)^2} -36 -2\left( {2{x^2} -13x -7} \right)\\ &= {\left( {2x} \right)^2} -2 \times 2x \times 5 + {5^2} -36 -4{x^2} + 26x + 14\\ &= 4{x^2} -20x + 25 -36 -4{x^2} + 26x + 14\\ &= 6x + 3\\ &= 3\left( {2x + 1} \right) \end{aligned}$$
3. On a :
   $$B -2A = 3\left( {2x + 1} \right)$$
   Alors :
   $$\begin{aligned} 2A &= B + 3\left( {2x + 1} \right)\\ 2A &= \left( {{{2x -11}}} \right)\left( {{{2x + 1}}} \right) + 3\left( {2x + 1} \right)\\ 2A &= \left( {{{2x + 1}}} \right)\left( {2x -11 + 3} \right)\\ 2A &= \left( {{{2x + 1}}} \right)\left( {2x -8} \right)\\ 2A &= 2\left( {{{2x + 1}}} \right)\left( {x -4} \right) \end{aligned}$$
   D’où : $$A= \left( {{{2x + 1}}} \right)\left( {x -4} \right).$$