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a. \(x^4-4x^3+8x^2-16x+16\)
\(=x^4-2x^3-2x^3+4x^2+4x^2-8x-8x+16\)
\(=x^3\left(x-2\right)-2x^2\left(x-2\right)+4x\left(x-2\right)-8\left(x-2\right)\)
\(=\left(x-2\right)\left(x^3-2x^2+4x-8\right)\)
\(=\left(x-2\right)\left[x^2\left(x-2\right)+4\left(x-2\right)\right]\)
\(=\left(x-2\right)^2\left(x^2+4\right)\)
b. \(x^4-25x^2+20x-4\)
\(=x^4+5x^3-5x^3-25x^2+2x^2-2x^2+10x+10x-4\)
\(=\left(x^4+5x^3-2x^2\right)-\left(5x^3+25x^2-10x\right)+\left(2x^2+10x-4\right)\)
\(=x^2\left(x^2+5x-2\right)-5x\left(x^2+5x-2\right)+2\left(x^2+5x-2\right)\)
\(=\left(x^2+5x-2\right)\left(x^2-5x+2\right)\)
a)\(x^6-y^6=\left(x^3-y^3\right)\left(x^3+y^3\right)\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(x^2+9x-y^2-9=saide\)
\(a^2-14a-9b^2+49=\left(a-3b-7\right)\left(a+3b-7\right)\)
\(9-a^2-2ab-b^2=-\left(a+b-3\right)\left(a+b+3\right)\)
\(x^2-10x-y^2+10y=\left(x-y\right)\left(x+y-10\right)\)
\(a^2+2ab+b^2-81=\left(a+b-9\right)\left(a+b+9\right)\)
\(x^2-y^2+5x-5y=\left(x-y\right)\left(x+y+5\right)\)
a) \(\sqrt{x^2}=7\)
\(\Leftrightarrow\left|x\right|=7\)
\(\Leftrightarrow\orbr{\begin{cases}x=7\\x=-7\end{cases}}\)
b) \(\sqrt{\left(x-2020\right)^2}=10\)
\(\Leftrightarrow\left|x-2020\right|=10\)
\(\Leftrightarrow\orbr{\begin{cases}x-2020=10\\x-2020=-10\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2030\\x=2010\end{cases}}\)
c) đk: \(x\ge2\)
\(\sqrt{4}-\left(x-2\right)+3\sqrt{16x-32}=8\)
\(\Leftrightarrow2-x+2+12\sqrt{x-2}=8\)
\(\Leftrightarrow12\sqrt{x-2}=x+4\)
\(\Leftrightarrow144\left(x-2\right)=\left(x+4\right)^2\)
\(\Leftrightarrow x^2-136x+304=0\)
\(\Leftrightarrow\orbr{\begin{cases}x_1=133,726...\\x_2=2,273...\end{cases}}\)
d) đk: \(x\ge-1\)
\(\sqrt{25x+25}-2\sqrt{64x+64}=7\)
\(\Leftrightarrow5\sqrt{x+1}-16\sqrt{x+1}=7\)
\(\Leftrightarrow-11\sqrt{x+1}=7\)
Mà \(-11\sqrt{x+1}\le0< 7\left(\forall x\right)\)
=> pt vô nghiệm
a.
ĐKXĐ: $x\geq 0$
PT $\Leftrightarrow 6\sqrt{2x}-4\sqrt{2x}+5\sqrt{2x}=21$
$\Leftrightarrow 7\sqrt{2x}=21$
$\Leftrightarrow \sqrt{2x}=3$
$\Leftrightarrow 2x=9$
$\Leftrightarrow x=\frac{9}{2}$ (tm)
b.
ĐKXĐ: $x\geq -2$
PT $\Leftrightarrow \sqrt{25(x+2)}+3\sqrt{4(x+2)}-2\sqrt{16(x+2)}=15$
$\Leftrightarrow 5\sqrt{x+2}+6\sqrt{x+2}-8\sqrt{x+2}=15$
$\Leftrightarrow 3\sqrt{x+2}=15$
$\Leftrightarrow \sqrt{x+2}=5$
$\Leftrightarrow x+2=25$
$\Leftrightarrow x=23$ (tm)
c.
$\sqrt{(x-2)^2}=12$
$\Leftrightarrow |x-2|=12$
$\Leftrightarrow x-2=12$ hoặc $x-2=-12$
$\Leftrightarrow x=14$ hoặc $x=-10$
e.
PT $\Leftrightarrow |2x-1|-x=3$
Nếu $x\geq \frac{1}{2}$ thì $2x-1-x=3$
$\Leftrightarrow x=4$ (tm)
Nếu $x< \frac{1}{2}$ thì $1-2x-x=3$
$\Leftrightarrow x=\frac{-2}{3}$ (tm)
a/ ĐKXĐ: ....
\(\Leftrightarrow2x^2+2x+4+2x-4=5\sqrt{\left(x-2\right)\left(x^2+x+2\right)}\)
\(\Leftrightarrow2\left(x^2+x+2\right)+2\left(x-2\right)=5\sqrt{\left(x-2\right)\left(x^2+x+4\right)}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+x+2}=a\\\sqrt{x-2}=b\end{matrix}\right.\)
\(\Leftrightarrow2a^2+2b^2=5ab\)
\(\Leftrightarrow\left(a-2b\right)\left(2a-b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=2b\\2a=b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+x+2}=2\sqrt{x-2}\\2\sqrt{x^2+x+2}=\sqrt{x-2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x+2=4\left(x-2\right)\\4\left(x^2+x+2\right)=x-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-3x+10=0\\4x^2+3x+10=0\end{matrix}\right.\)
Phương trình vô nghiệm
b/ ĐKXĐ: ....
\(\Leftrightarrow2x^2-x+1=\sqrt{4x^4+4x^2+1-4x^2}\)
\(\Leftrightarrow2x^2-x+1=\sqrt{\left(2x^2+1\right)^2-\left(2x\right)^2}\)
\(\Leftrightarrow2x^2-x+1=\sqrt{\left(2x^2-2x+1\right)\left(2x^2+2x+1\right)}\)
\(\Leftrightarrow\frac{3}{4}\left(2x^2-2x+1\right)+\frac{1}{4}\left(2x^2+2x+1\right)=\sqrt{\left(2x^2-2x+1\right)\left(2x^2+2x+1\right)}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{2x^2-2x+1}=a\\\sqrt{2x^2+2x+1}=b\end{matrix}\right.\)
\(\Leftrightarrow3a^2+b^2=4ab\Leftrightarrow3a^2-4ab+b^2=0\)
\(\Leftrightarrow\left(a-b\right)\left(3a-b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\\3a=b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2x^2-2x+1}=\sqrt{2x^2+2x+1}\\3\sqrt{2x^2-2x+1}=\sqrt{2x^2+2x+1}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x^2-2x+1=2x^2+2x+1\\9\left(2x^2-2x+1\right)=2x^2+2x+1\end{matrix}\right.\)
lộn toán lớp 8 ko phải toan lop 9