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\(3\left(x-2\right)+4\left(x-1\right)=25\)
\(\Leftrightarrow3x-6+4x-4=25\)
\(\Leftrightarrow7x=35\)
\(\Leftrightarrow x=5\)
\(\left(5x-3\right)\left(x-2\right)=\left(x-1\right)\left(x-2\right)\)
\(\Leftrightarrow\left(5x-3\right)\left(x-2\right)-\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(5x-3-x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(4x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\4x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-1}{2}\end{matrix}\right.\)
a) \(\Rightarrow5x^2-15x=5x^2-x-10x+2-5\)
\(\Rightarrow5x^2-15x-5x^2+x+10x=2-5\)
\(\Rightarrow-4x=-3\)
\(\Rightarrow x=\frac{3}{4}\)
Vậy \(x=\frac{3}{4}\)
b) \(\Rightarrow x^2-4x-5x+20-x^2+2x-x+2=7\)
\(\Rightarrow x^2-4x-5x-x^2+2x-x=7-20-2\)
\(\Rightarrow-8x=-15\)
\(\Rightarrow x=\frac{15}{8}\)
Vậy \(x=\frac{15}{8}\)
c) \(\Rightarrow3x^2-6x-4x+8=3x^2-27x-3\)
\(\Rightarrow3x^2-6x-4x-3x^2+27x=-3-8\)
\(\Rightarrow17x=-11\)
\(\Rightarrow x=\frac{-11}{17}\)
Vậy \(x=\frac{-11}{17}\)
Chúc bạn học tốt.
a)
<=> 10x - 35 + 16x - 10 = 5
<=> 10x + 16x = 5 + 35 + 10
<=> 26x = 50
<=> x = 50/26 = 25/13
a, ( 2x - 3 )2- (2x + 1)2 = -3
4x2-12x+9-4x2+4x-1=-3
-8x-1=-3
-8x=-2
x=\(\frac{1}{4}\)
b, (5x - 1) 2 - (5x + 4)(5x - 4) = 7
25x2-10x+1-25x2+16=7
-10x+17=7
-10x=-10
x=1
c, ( x- 5)2 + (x-3)(x+3) - 2(x + 1)2=0
x2-10x+25+x2-9-2x2-4x-2=0
-14x+14=0
-14(x-1)=0
=>x-1=0
x=1
a) \(\left(2x-3\right)^2-\left(2x+1\right)^2=-3\)
\(\Leftrightarrow4x^2-12x+9-4x^2-4x-1=-3\)
\(\Leftrightarrow-16x+8=-3\)
\(\Leftrightarrow-16x=-11\)
\(\Leftrightarrow x=\frac{11}{16}\)
b)\(\left(5x-1\right)^2-\left(5x+4\right)\left(5x-4\right)=7\)
\(\Leftrightarrow25x^2-10x+1-25x^2+16=7\)
\(\Leftrightarrow-10x+17=7\)
\(\Leftrightarrow-10x=-10\)
\(\Leftrightarrow x=1\)
c)\(\left(x-5\right)^2+\left(x-3\right)\left(x+3\right)-2\left(x+1\right)^2=0\)
\(\Leftrightarrow x^2-10x+25+x^2-9-2\left(x^2+2x+1\right)=0\)
\(\Leftrightarrow2x^2-10x-16-2x^2-4x-2=0\)
\(\Leftrightarrow-14x-18=0\)
\(\Leftrightarrow-14x=18\)
\(\Leftrightarrow x=-\frac{9}{7}\)
#H
\(\left(8-5x\right)\left(x+2\right)+4\left(x-2\right)\left(x+1\right)+2\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow8x+16-5x^2-10x+4x^2+4x-8x-8+2x^2-8=0\)
\(\Leftrightarrow x^2-6x=0\Leftrightarrow x\left(x-6\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x-6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=6\end{cases}}}\)
Vậy S = { 0, 6}
|x+4|-2|x-1|=5x => x + 4 - 2 ( x - 1) = 5x => x + 4 - 2x + 2 = 5x => 6x = 6 => x=1