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9 tháng 2 2017
a) Đặt x^2+2x+2=t
\(\frac{4}{t-1}+\frac{3}{t+1}=\frac{3}{2}\Leftrightarrow\frac{4t+4+3t-3}{t^2-1}=\frac{7t+1}{t^2-1}=\frac{3}{2}\)
\(\Leftrightarrow14t+2=3t^2-3\Leftrightarrow3t^2-14t-5=3t\left(t-5\right)+t-5=0\)\(\Leftrightarrow\left(t-5\right)\left(3t+1\right)=0\Rightarrow\left[\begin{matrix}t=5\\t=-\frac{1}{3}\left(loai\right)\end{matrix}\right.\)
Với t=5 ta có (x+1)^2=4\(\Rightarrow\left[\begin{matrix}x+1=2\\x+1=-2\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=1\\x=-3\end{matrix}\right.\)
2 tháng 9 2022
1: \(\Leftrightarrow-4x^2+3x-4x^2+8x=10\)
=>-8x^2+11x-10=0
=>\(x\in\varnothing\)
2: \(\Leftrightarrow5x^2-15x+5+x-5x^2=x-2\)
=>-14x+5=x-2
=>-15x=-7
=>x=7/15
3: \(\Leftrightarrow12x^2-12x^2+20x=10x-17\)
=>10x=-17
=>x=-17/10
4: \(\Leftrightarrow4x^2-2x+3-4x^2+20x=7x-3\)
=>18x+3=7x-3
=>11x=-6
=>x=-6/11
5: \(\Leftrightarrow-3x+15+5x-5+3x^2=4-x\)
\(\Leftrightarrow3x^2+2x+10-4+x=0\)
=>3x^2+3x+6=0
hay \(x\in\varnothing\)
NT
1
QN
4
1 tháng 9 2018
a) \(\left(5x-2\right)\left(5x+2\right)-\left(5x+3\right)\left(5x-4\right)=0\)
\(\Leftrightarrow5x+8=8\)
\(\Leftrightarrow5x=8-8\)
\(\Leftrightarrow x=5.0\)
\(\Leftrightarrow x=0\)
b)
9 tháng 7 2018
A. \(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left(x^2+3x+2x+6\right)-\left(x^2+5x-2x-10\right)=0\)
\(\Leftrightarrow x^2+3x+2x+6-x^2-5x+2x+10=0\)
\(\Leftrightarrow x^2+3x+2x-x^2-5x+2x=-6-10\)
\(\Leftrightarrow2x=-16\)
\(\Leftrightarrow x=-8\) .Vậy \(S=\left\{-8\right\}\)
B. \(\left(2x+3\right)\left(x-4\right)+\left(x-5\right)\left(x-2\right)=\left(3x+5\right)\left(x-4\right)\)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x-5x+10=3x^2-12x+5x-20\)
\(\Leftrightarrow2x^2-8x+3x+x^2-2x-5x-3x^2+12x-5x=12-10-20\)
\(\Leftrightarrow-5x=-18\)
\(\Leftrightarrow x=\dfrac{18}{5}\) . Vậy \(S=\left\{\dfrac{18}{5}\right\}\)
C. \(\left(8-4x\right)\left(x+2\right)+4\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow8x+16-4x^2-8x+4\left(x^2+x-2x-2\right)=0\)
\(\Leftrightarrow8x+16-4x^2-8x+4x^2+4x-8x-8=0\)
\(\Leftrightarrow8x-4x^2-8x+4x^2+4x-8x=-16+8\)
\(\Leftrightarrow-4x=-8\)
\(\Leftrightarrow x=2\) . Vậy \(S=\left\{2\right\}\)
D. \(\left(2x-3\right)\left(8x+2\right)=\left(4x+1\right)\left(4x-1\right)-3\)
\(\Leftrightarrow16x^2+4x-24x-6=16x^2+1^2-3\)
\(\Leftrightarrow16x^2+4x-24x-16x^2=6+1-3\)
\(\Leftrightarrow-20x=4\)
\(\Leftrightarrow x=-\dfrac{1}{5}\) . Vậy \(S=\left\{-\dfrac{1}{5}\right\}\)
9 tháng 7 2018
a)(x+2)(x+3)-(x-2)(x+5)=0
\(\Leftrightarrow x^2+3x+2x+6-x^2-5x+2x+10=0\)
<=>2x=-16
<=>x=-8
b)(2x+3)(x-4)+(x-5)(x-2)=(3x-5)(x-4)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x-5x+10=3x^2-12x-5x+20\)
\(\Leftrightarrow3x^2-12x-2=3x^2-17x+20\)
\(\Leftrightarrow5x=22\Leftrightarrow x=\dfrac{22}{5}\)
c)(8-4x)(x+2)+4(x-2)(x+1)=0
\(\Leftrightarrow8x+16-4x^2-8x+4x^2+4x-8x-8=0\)
\(\Leftrightarrow-4x=-8\Leftrightarrow x=2\)
d)(2x-3)(8x+2)=(4x+1)(4x-1)-3
\(\Leftrightarrow16x^2+4x-24x-6=16x^2-4x+4x-1-3\)
\(\Leftrightarrow-20x=-2\Leftrightarrow x=\dfrac{-1}{10}\)
QN
TÌM X biết:
a. (5x - 2)(5x + 2) - (5x + 3)(5x - 4) = 8
b. (4x - 3)( 4x + 2) + (4x + 5)(1 - 4x) =2.52
1
KB
1 tháng 9 2018
a ) \(\left(5x-2\right)\left(5x+2\right)-\left(5x+3\right)\left(5x-4\right)=8\)
\(\Leftrightarrow\left(5x\right)^2-4-\left(25x^2+15x-20x-12\right)=8\)
\(\Leftrightarrow25x^2-4-25x^2-15x+20x+12=8\)
\(\Leftrightarrow5x+8=8\)
\(\Leftrightarrow5x=0\)
\(\Leftrightarrow x=0\)
Vậy \(x=0\)
b ) \(\left(4x-3\right)\left(4x+2\right)+\left(4x+5\right)\left(1-4x\right)=2.5^2\)
\(\Leftrightarrow16x^2-12x+8x-6+4x+5-16x^2-20x=50\)
\(\Leftrightarrow-20x-1=50\)
\(\Leftrightarrow-20x=51\)
\(\Leftrightarrow x=-\dfrac{51}{20}\)
Vậy \(x=-\dfrac{51}{20}\)
\(PT\Leftrightarrow x^5-1=4\left(x^4+x^3+x^2+x+1\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x^4+x^3+x^2+x+1\right)=4\left(x^4+x^3+x^2+x+1\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=4\\x^4+x^3+x^2+x+1=0\end{matrix}\right.\).
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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