tim x biet: (2x-3)^10x=(3-2x)^100
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HN
0
MT
1
8 tháng 12 2022
\(\Leftrightarrow\left(2x+3-2x+5\right)^2=x^2+6x+64\)
=>x^2+6x=0
=>x(x+6)=0
=>x=0 hoặc x=-6
MR
1
KT
22 tháng 7 2018
\(x^2-5x-4\left(x-5\right)=0\)
\(\Leftrightarrow\)\(x\left(x-5\right)-4\left(x-5\right)=0\)
\(\Leftrightarrow\)\(\left(x-5\right)\left(x-4\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x-5=0\\x-4=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=5\\x=4\end{cases}}\)
Vậy....
\(2x\left(x+6\right)=7x+42\)
\(\Leftrightarrow\)\(2x\left(x+6\right)-7x-42=0\)
\(\Leftrightarrow\)\(2x\left(x+6\right)-7\left(x+6\right)=0\)
\(\Leftrightarrow\)\(\left(x+6\right)\left(2x-7\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x+6=0\\2x-7=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-6\\x=\frac{7}{2}\end{cases}}\)
Vậy......
\(x^3-5x^2+x-5=0\)
\(\Leftrightarrow\)\(x^2\left(x-5\right)+\left(x-5\right)=0\)
\(\Leftrightarrow\)\(\left(x-5\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\)\(x-5=0\)
\(\Leftrightarrow\)\(x=5\)
\(x^4-2x^3+10x^2-20x=0\)
\(\Leftrightarrow\)\(x^3\left(x-2\right)+10x\left(x-2\right)=0\)
\(\Leftrightarrow\)\(x\left(x-2\right)\left(x^2+10\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
Vậy...
11 tháng 8 2017
\(\left|x-\dfrac{1}{2}\right|\left|2x-\dfrac{3}{4}\right|=2x-\dfrac{3}{4}\)
\(\left\{{}\begin{matrix}\left|x-\dfrac{1}{2}\right|\ge0\\\left|2x-\dfrac{3}{4}\right|\ge0\end{matrix}\right.\)
\(\Rightarrow2x-\dfrac{3}{4}\ge0\)
\(\Rightarrow\left|2x-\dfrac{3}{4}\right|=2x-\dfrac{3}{4}\)
\(\Rightarrow\left|x-\dfrac{1}{2}\right|=1\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=1\Rightarrow x=\dfrac{3}{2}\\x-\dfrac{1}{2}=-1\Rightarrow x=-\dfrac{1}{2}\end{matrix}\right.\)
\(2x-\dfrac{3}{4}\ge0\Rightarrow2x\ge\dfrac{3}{4}\Rightarrow x\ge\dfrac{3}{2}\)
Vậy xảy ra khi:
\(x=\dfrac{3}{2}\)
7 tháng 10 2021
a: Ta có: \(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)
\(\Leftrightarrow\left(4x+14-3x-9\right)\left(4x+14+3x+9\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(7x+23\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{23}{7}\end{matrix}\right.\)
c: Ta có: \(\left(x-3\right)^2-4=0\)
\(\Leftrightarrow\left(x-5\right)\cdot\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
AH
Akai Haruma
Giáo viên
8 tháng 10 2021
b.
PT $\Leftrightarrow (5x^2-2x+10)^2-(3x^2+10x-8)^2=0$
$\Leftrightarrow (5x^2-2x+10-3x^2-10x+8)(5x^2-2x+10+3x^2+10x-8)=0$
$\Leftrightarrow (2x^2-12x+18)(8x^2+8x+2)=0$
$\Leftrightarrow (x^2-6x+9)(4x^2+4x+1)=0$
$\Leftrightarrow (x-3)^2(2x+1)^2=0$
$\Leftrightarrow (x-3)(2x+1)=0$
$\Leftrightarrow x-3=0$ hoặc $2x+1=0$
$\Leftrightarrow x=3$ hoặc $x=-\frac{1}{2}$
d.
$x^2-2x=24$
$\Leftrightarrow x^2-2x-24=0$
$\Leftrightarrow (x+4)(x-6)=0$
$\Leftrightarrow x+4=0$ hoặc $x-6=0$
$\Leftrightarrow x=-4$ hoặc $x=6$
7 tháng 8 2021
\(\left(3-\sqrt{2x}\right)\left(2-3\sqrt{2x}\right)=6x-5\)ĐK : x>= 0
\(\Leftrightarrow6-9\sqrt{2x}-2\sqrt{2x}+6x=6x-5\)
\(\Leftrightarrow-11\sqrt{2x}+6x+6=6x-5\Leftrightarrow-11\sqrt{2x}=-11\)
\(\Leftrightarrow\sqrt{2x}=1\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\left(tm\right)\)
\(\left(2x-3\right)^{10x}=\left(3-2x\right)^{100}\)
\(\Leftrightarrow\left(2x-3\right)^{10x}=\left(2x-3\right)^{100}\)
\(\Leftrightarrow10x=100\Leftrightarrow x=100:10=10\)
\(\left(2x-3\right)^{10x}=\left(3-2x\right)^{100}\)
\(\Rightarrow\left(2x-3\right)^{10x}=\left(2x-3\right)^{100}\)
\(\Rightarrow10x\Leftrightarrow x=100:10=10\)
~Study well~ :)