/7x-2x/+7=2x. tìm x
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a) 5(2x -1) - 4(8 - 3x) = 7
<=> 10x - 5 - 32 + 12x = 7
<=> 22x = 44
<=> x =2
Vậy x = 2 là nghiệm phương trình
b) 7(2x - 5) - 5(7x - 2) + 2(5x - 7) = (x - 2) - (x + 4)
<=> 14x - 35 - 35x + 10 + 10x - 14 = x - 2 - x - 4
<=> -11x - 39 = - 6
<=> -11x = 33
<=> x = -3
Vậy x = -3 là nghiệm phương trình
\(a,10x-5-32+12x=7\)
\(22x=44\)
\(x=2\)
\(b,14x-35-35x+10+10x-14=x-2-x-4\)
\(-11x-39=-6\)
\(-11x=-33\)
\(x=3\)
a)2x(x-5)-2x2 = 20
\(VT=-10x\)
\(\Leftrightarrow-10x=20\)
\(\Leftrightarrow x=-2\)
b)5x(2x-7)+2x(8-5x)=5
\(VT=-19x\)
\(\Leftrightarrow-19x=5\)
\(\Leftrightarrow x=-\frac{5}{19}\)
c)4x(7x-5)-7x(4x-2)= -12
\(VT=-6x\)
\(\Leftrightarrow-6x=-12\)
\(\Leftrightarrow x=2\)
7x( 2 - 3x ) + x2( 2x + 1 ) - 2x2( x - 2 ) + 2x( 8x - 7 )
= 14x - 21x2 + 2x3 + x2 - 2x3 + 4x2 + 16x2 - 14x
= ( 2x3 - 2x3 ) + ( -21x2 + x2 + 4x2 + 16x2 ) + ( 14x - 14x )
= 0
a) ( 2x - 3 ) - ( x - 5 ) = ( x + 7 ) - ( x + 2 )
<=> 2x - 3 - x + 5 = x + 7 - x - 2
<=> x = 3
b)(7x-5)-(6x+4)=(2x+3)-(2x+1)
<=> 7x - 5 - 6x - 4 = 2x + 3 - 2x - 1
<=> x = 11
c)(9x-3)-(8x+5)=(3x+2)
<=> 9x - 3 - 8x - 5 = 3x + 2
<=> -2x = 10
<=> x = -5
d)(x+7)-(2x+3)=(3x+5)-(2x+4)
<=> x + 7 - 2x - 3 = 3x + 5 - 2x - 4
<=> -2x = -3
<=> x = 3/2
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a) \(7x^2-16x=2x^3-56\)
\(\Leftrightarrow\)\(2x^3-7x^2+16x-56=0\)
\(\Leftrightarrow\)\(2x\left(x^2+8\right)-7\left(x^2+8\right)=0\)
\(\Leftrightarrow\)\(\left(2x-7\right)\left(x^2+8\right)=0\)
\(\Leftrightarrow\)\(2x-7=0\)
\(\Leftrightarrow\)\(x=3,5\)
Vậy...
b) \(x^7+x^3+2x^5+2x=0\)
\(\Leftrightarrow\)\(x.\left(x^6+x^2+2x^4+2\right)=0\)
\(\Leftrightarrow\)\(x\left(x^2+2\right)\left(x^4+1\right)=0\)
\(\Leftrightarrow\)\(x=0\)
Vậy...
c) \(\left(2x+1\right)x-5\left(x+\frac{1}{2}\right)=0\)
\(\Leftrightarrow\)\(2x\left(x+\frac{1}{2}\right)-5\left(x+\frac{1}{2}\right)=0\)
\(\Leftrightarrow\)\(\left(2x-5\right)\left(x+\frac{1}{2}\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}2x-5=0\\x+\frac{1}{2}=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=2,5\\x=-0,5\end{cases}}\)
Vậy...
a) \(7x\left(2x-3\right)-\left(4x^2-9\right)=0\Rightarrow7x\left(2x-3\right)-\left(2x-3\right)\left(2x+3\right)=0\Rightarrow\left(2x-3\right)\left(7x-2x+3\right)=0\Rightarrow\left[{}\begin{matrix}2x-3=0\\5x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{5}\end{matrix}\right.\)
b) \(\left(2x-7\right).\left(x-2\right)\left(x^2-4\right)=0\Rightarrow\left(2x-7\right)\left(x-2\right)^2\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}2x-7=0\\\left(x-2\right)^2=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=2\\x=-2\end{matrix}\right.\)
c)\(\left(9x^2-25\right)-\left(6x-10\right)=0\Rightarrow\left(3x-5\right)\left(3x+5\right)-2\left(3x-5\right)=0\Rightarrow\left(3x-5\right)\left(3x+5-2\right)=0\Rightarrow\left[{}\begin{matrix}3x-5=0\\3x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=1\end{matrix}\right.\)
a: Ta có: \(7x\left(2x-3\right)-\left(4x^2-9\right)=0\)
\(\Leftrightarrow7x\left(2x-3\right)-\left(2x-3\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(5x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{5}\end{matrix}\right.\)
b: Ta có: \(\left(2x-7\right)\left(x-2\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(2x-7\right)\left(x-2\right)^2\cdot\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=2\\x=-2\end{matrix}\right.\)
c: Ta có: \(\left(9x^2-25\right)-\left(6x-10\right)=0\)
\(\Leftrightarrow\left(3x-5\right)\left(3x+5-2\right)=0\)
\(\Leftrightarrow\left(3x-5\right)\left(3x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-1\end{matrix}\right.\)
a) \(\dfrac{3x-4}{2x+5}=\dfrac{3x+7}{2x-20}\left(đk:x\ne-\dfrac{5}{2},x\ne10\right)\)
\(\Rightarrow\left(3x-4\right)\left(2x-20\right)=\left(3x+7\right)\left(2x+5\right)\)
\(\Rightarrow6x^2-68x+80=6x^2+29x+35\)
\(\Rightarrow97x=45\Rightarrow x=\dfrac{45}{97}\)
b) \(\dfrac{10x-5}{7x+2}=\dfrac{50x+10}{35x-29}\left(đk:x\ne-\dfrac{2}{7},x\ne\dfrac{29}{35}\right)\)
\(\Rightarrow\left(10x-5\right)\left(35x-29\right)=\left(50x+10\right)\left(7x+2\right)\)
\(\Rightarrow350x^2-465x+145=350x^2+170x+20\)
\(\Rightarrow635x=125\Rightarrow x=\dfrac{25}{127}\)
a,(2x-3)^2-(2x-3)(x+1)=0
(2x-3)(2x-3-x-1)=0
(2x-3)(x-4)=0
2x-3=0 or x-4=0
x=3/2 or x=4