|4x-3|=5
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\(PT\Leftrightarrow x^5-1=4\left(x^4+x^3+x^2+x+1\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x^4+x^3+x^2+x+1\right)=4\left(x^4+x^3+x^2+x+1\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=4\\x^4+x^3+x^2+x+1=0\end{matrix}\right.\).
Nếu \(x^4+x^3+x^2+x+1=0\Rightarrow\left(x-1\right)\left(x^4+x^3+x^2+x+1\right)=0\Leftrightarrow x^5-1=0\Leftrightarrow x^5=1\Leftrightarrow x=1\). Thử lại ta thấy không thoả mãn.
Do đó ta có \(x-1=4\Leftrightarrow x=5\).
Vậy...
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a: A=(4x+5)^2-2*(4x+5)(4x-5)+(4x-5)^2
=(4x+5-4x+5)^2
=10^2=100
b: B=(3x-2)^2*(3x+2)^2-2(2x+3)(2x-3)
=(9x^2-4)^2-2(4x^2-9)
=81x^4-72x^2+16-8x^2+18
=81x^4-80x^2+34
\(a,A=\left(4x-5\right)^2+\left(4x+5\right)^2+2\left(5+4x\right)\left(5-4x\right)\)
\(=\left(5-4x\right)^2 +2\left(5-4x\right)\left(4x+5\right)+\left(4x+5\right)^2\)
\(=\left(5-4x+4x+5\right)^2\)
\(=10^2\)
\(=100\)
\(b,B=\left(3x-2\right)^2\left(3x+2\right)^2-2\left(2x+3\right)\left(2x-3\right)\)
\(=\left(9x^2-4\right)^2-2\left(4x^2-9\right)\)
\(=81x^4-72x^2+16-8x^2+18\)
\(=81x^4-80x^2+34\)
#\(Urushi\)
(5-4x).(5+4x)-(4x-3)2
= 25-16x2-(16x2-24x+9)
= 25-16x2-16x2+24x-9
=-32x2+ 24x+16
2: 12-10x=25-30x
=>20x=13
=>x=13/20
3: \(3\left(2x+3\right)-2\left(4x-5\right)=10x+21\)
=>6x+9-8x+10=10x+21
=>10x+21=-2x+19
=>12x=-2
=>x=-1/6
4: \(\Leftrightarrow25x-15-6x+12=11-5x\)
=>19x-3=11-5x
=>24x=14
=>x=7/12
5: \(\Leftrightarrow8-12x-5+10x=4-6x\)
=>4-6x=-2x+3
=>-4x=-1
=>x=1/4
6: \(\Leftrightarrow32x-24-6+9x=13-40x\)
=>41x-30=13-40x
=>81x=43
=>x=43/81
7: \(\Leftrightarrow10x-5+20x=5x-11\)
=>30x-5=5x-11
=>25x=-6
=>x=-6/25
\(\left|2+3x\right|=\left|4x-3\right|\)
\(\Leftrightarrow\orbr{\begin{cases}2+3x=4x-3\\2+3x=3-4x\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\\x=\frac{1}{7}\end{cases}}\)
Vậy \(x\in\left\{\frac{1}{7};5\right\}\)
\(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=1-4x\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}\)
Vậy \(x\in\left\{\frac{1}{11};\frac{3}{5}\right\}\)
\(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
\(\Leftrightarrow\left|\frac{5}{4}x-\frac{7}{2}\right|=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
Giải tiếp tương tự
Sau đó giải tiếp câu còn lại
\(a)PT\Leftrightarrow4x^2-9-4x^2+20x+3x=0.\\ \Leftrightarrow23x=9.\\ \Leftrightarrow x=\dfrac{9}{23}.\\ b)PT\Leftrightarrow\left(2x+1\right)\left(4x-3\right)-\left(2x+1\right)\left(2x-1\right)=0.\\\Leftrightarrow\left(2x+1\right)\left(4x-3-2x+1\right)=0.\\ \Leftrightarrow\left(2x+1\right)\left(2x-2\right)=0.\\ \Leftrightarrow\left(2x+1\right)\left(x-1\right)=0. \)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{2}.\\x=1.\end{matrix}\right.\)
|4x-3|=5
x= 2