Bài 1: Tìm x
a) \(\left(5-2x\right)^2-16=0\)
b) \(x^2-4x=29\)
c) \(\left(x-3\right)^3-\left(x-3\right).\left(x^2+3x+9\right)+9.\left(x+1\right)^2=15\)
d) \(2.\left(x-5\right).\left(x+5\right)-\left(x+2\right).\left(2x-3\right)+x.\left(x^2-8\right)=\left(x+1\right).\left(x^2-x+1\right)\)
Bài 2: Rút gọn
a) \(\left(x^2+x+1\right).\left(x^2-x+1\right).\left(x^2-1\right)\)
b) \(\left(a+b-c\right)^2+\left(a-b+c\right)^2-2.\left(b-c\right)^2\)
c) \(\left(a+b+c\right)^2-\left(a+b\right)^2-\left(a+c\right)^2-\left(b+c\right)^2\)
d) \(\left(a+b+c\right)^2+\left(a-b-c\right)^2+\left(b-c-a\right)^2+\left(c-a-b\right)^2\)
1. a) $(5-2x)^2-16=0$
$=>(5-2x)^2-4^2=0$
$=>(5-2x-4)(5-2x+4)=0$
$=>(1-2x)(9-2x)=0$
\(=>\left[{}\begin{matrix}1-2x=0=>x=0,5\\9-2x=0=>x=4,5\end{matrix}\right.\)
b) $x^2-4x=29$
$=>x^2-4x-29=0$
$=>(x^2-4x+4)-33=0$
$=>(x-2)^2-(\sqrt{33})^2=0$
$=>(x-2-\sqrt{33})(x-2+\sqrt{33})=0$
\(=>\left[{}\begin{matrix}x-2-\sqrt{33}=0=>x=\sqrt{33}+2\\x-2+\sqrt{33}=0=>x=2-\sqrt{33}\end{matrix}\right.\)
Bài 1:
a) \(\left(5-2x\right)^2-16=0\) (1)
\(\Leftrightarrow\left(5-2x\right)^2=16\)
\(\Leftrightarrow5-2x=\pm4\)
\(\Leftrightarrow\left[{}\begin{matrix}5-2x=4\\5-2x=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{9}{2}\end{matrix}\right.\)
Vậy tập nghiệm phương trình (1) là \(S=\left\{\dfrac{1}{2};\dfrac{9}{2}\right\}\)
b) \(x^2-4x=29\) (2)
\(\Leftrightarrow x^2-4x-29=0\)
\(\Leftrightarrow x=\dfrac{4\pm2\sqrt{33}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4+2\sqrt{33}}{2}\\x=\dfrac{4-2\sqrt{33}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2+\sqrt{33}\\x=2-\sqrt{33}\end{matrix}\right.\)
Vậy tập nghiệm phương trình (2) là \(S=\left\{2-\sqrt{33};2+\sqrt{33}\right\}\)
c) \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+9\left(x+1\right)^2=15\) (3)
\(\Leftrightarrow x^3-9x^2+27x-27-\left(x^3-27\right)+9\left(x^2+2x+1\right)=15\)
\(\Leftrightarrow x^3-9x^2+27x-27-\left(x^3-27\right)+9x^2+18x+9=15\)
\(\Leftrightarrow x^3+27x-27-x^3+27+18x+9=15\)
\(\Leftrightarrow45x+9=15\)
\(\Leftrightarrow45x=15-9\)
\(\Leftrightarrow45x=6\)
\(\Leftrightarrow x=\dfrac{2}{15}\)
Vậy tập nghiệm phương trình (3) là \(S=\left\{\dfrac{2}{15}\right\}\)
d) \(2\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(2x-3\right)+x\left(x^2+8\right)=\left(x+1\right)\left(x^2-x+1\right)\)(4)
\(\Leftrightarrow2\left(x^2-25\right)-\left(2x^2-3x+4x-6\right)+x^3-8x=x^3+1\)
\(\Leftrightarrow2x^2-50-\left(2x^2+x-6\right)+x^3-8x=x^3+1\)
\(\Leftrightarrow2x^2-50-2x^2-x+6-8x=1\)
\(\Leftrightarrow-44-9x=1\)
\(\Leftrightarrow-9x=1+45\)
\(\Leftrightarrow-9x=45\)
\(\Leftrightarrow x=-5\)
Vậy tập nghiệm phương trình (4) là \(S=\left\{-5\right\}\)
Bài 2:
a) \(\left(x^2+x+1\right)\left(x^2-x+1\right)\left(x^2-1\right)\)
\(=\left(x^4-x^3+x^2+x^3-x^2+x+x^2-x+1\right)\left(x^2-1\right)\)
\(=x^6-x^4+x^4-x^2+x^2-1\)
\(=x^6-1\)
b) \(\left(a+b-c\right)^2+\left(a-b+c\right)^2-2\left(b-c\right)^2\)
\(=a^2+b^2+\left(-c\right)^2+2ab-2ac+a^2+\left(-b\right)^2+c^2-2ab+2ac-2bc-2\left(b^2+2bc+c^2\right)\)
\(=a^2+b^2+\left(-c\right)^2-2bc+a^2+\left(-b\right)^2+c^2-2bc-2b^2-4bc-2c^2\)
\(=2a^2-b^2+c^2-8bc+\left(-b\right)^2-c^2\)
\(=2a^2-b^2-8bc+b^2\)
\(=2a^2-8bc\)
c) \(\left(a+b+c\right)^2-\left(a+b\right)^2-\left(a+c\right)^2-\left(b+c\right)^2\)
\(=a^2+b^2+c^2+2ab+2ac+2bc-\left(a^2+2ab+b^2\right)-\left(a^2+2ac+b^2\right)-\left(b^2+2bc+c^2\right)\)
\(=a^2+b^2+c^2+2ab+2ac+2bc-a^2-2ab-b^2-a^2-2ac-c^2-b^2-2bc-c^2\)
\(=-a^2-b^2-c^2\)
d) \(\left(a+b+c\right)^2+\left(a-b-c\right)^2+\left(b-c-a\right)^2+\left(c-a-b\right)^2\)
\(=a^2+b^2+c^2+2ab+2ac+2bc+a^2+b^2+c^2-2ab-2ac+2bc+\\ b^2+c^2+a^2-2bc-2ab+2ac+c^2+a^2+b^2-2ac-2bc+2ab\)
\(=2a^2+2b^2+2c^2+2\cdot\left(-b\right)^2+2\cdot\left(-c\right)^2+2\cdot\left(-a\right)^2\)
\(=2a^2+2b^2+2c^2+2b^2+2c^2+2a^2\)
\(=4a^2+4b^2+4c^2\)
Bài 1:
a) \(\left(5-2x\right)^2-16=0\)
\(\Leftrightarrow\left(5-2x\right)^2=\left(\pm4\right)^2\)
\(\Leftrightarrow\left\{{}\begin{matrix}5-2x=-4\\5-2x=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=4,5\\x=0,5\end{matrix}\right.\)
Vậy x=4,5 hoặc x=0,5
b) \(x^2-4x=29\)
\(\Leftrightarrow x^2-4x-29=0\)
\(\Leftrightarrow x^2-4x+4-36=0\)
\(\Leftrightarrow\left(x-2\right)^2-36=0\)
\(\Leftrightarrow\left(x-2\right)^2=\left(\pm6\right)^2\)
\(\Rightarrow\left\{{}\begin{matrix}x-2=6\\x-2=-6\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=8\\x=-4\end{matrix}\right.\)
Vậy x=8 hoặc x=-4
