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a. \(x\left(x^2-25\right)-\left(x^3-2x^2+4x+2x^2-4x+8\right)=17\)
\(x^3-25x-\left(x^3+8\right)=17\)
\(x^3-25x-x^3-8=17\)
\(-25x=25\)
\(x=-1\)
c. \(6x^2-\left(6x^2-4x+15x-10\right)=7\)
\(6x^2-6x^2-11x+10=7\)
\(-11x=-3\)
\(x=\frac{3}{11}\)
Rút gọn hết ta được :
a/ 41x - 17 = -21
=> 41x = -4 => x = 4/41
b/ 34x - 17 = 0
=> 34x = 17
=> x = 17/34 = 1/2
c/ 19x + 56 = 52
=> 19x = -4
=> x = -4/19
d/ 20x2 - 16x - 34 = 10x2 + 3x - 34
=> 10x2 - 19x = 0
=> x(10x - 19) = 0
=> x = 0
hoặc 10x - 19 = 0 => 10x = 19 => x = 19/10
Vậy x = 0 ; x = 19/10
Rút gọn hết ta được :
a/ 41x - 17 = -21
=> 41x = -4 => x = 4/41
b/ 34x - 17 = 0
=> 34x = 17
=> x = 17/34 = 1/2
c/ 19x + 56 = 52
=> 19x = -4
=> x = -4/19
d/ 20x 2 - 16x - 34 = 10x 2 + 3x - 34
=> 10x 2 - 19x = 0
=> x(10x - 19) = 0
=> x = 0 hoặc 10x - 19 = 0
=> 10x = 19
=> x = 19/10
Vậy x = 0 ; x = 19/10
a) \(\left(x+3\right)\left(2x-1\right)-\left(x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow2x^2+5x-3-x^2+2x+3=0\)
\(\Leftrightarrow x^2+7x=0\Leftrightarrow x\left(x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-7\end{matrix}\right.\)
b) \(\left(x+4\right)\left(2x-3\right)-3\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow2x^2+5x-12-3x^2+12=0\)
\(\Leftrightarrow x^2-5x=0\Leftrightarrow x\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
a) \(x\left(x-4\right)-\left(x^2-8\right)=0\)
\(\Leftrightarrow x^2-4x-x^2+8=0\)
\(\Leftrightarrow-4\left(x-2\right)=0\)
\(\Leftrightarrow x-2=0\)
\(\Leftrightarrow x=2\)
b) \(\left(3x+2\right)\left(x-1\right)-3\left(x+1\right)\left(x-2\right)=4\)
\(\Leftrightarrow3x^2-3x+2x-2-3\left(x^2-2x+x-2\right)=0\)
\(\Leftrightarrow3x^2-3x+2x-2-3x^2+6x-3x+6=0\)
\(\Leftrightarrow2x=-4\)
\(\Leftrightarrow x=-2\)
c) \(x\left(x+5\right)\left(x-5\right)-\left(x+2\right)\left(x^2-2x+4\right)=17\)
\(\Leftrightarrow x\left(x^2-25\right)-\left(x^3+8\right)=17\)
\(\Leftrightarrow x^3-25x-x^3-8=17\)
\(\Leftrightarrow-25x=25\)
\(\Leftrightarrow x=-1\)
a) x(x - 4 ) - ( x^2 - 8 ) = 0
=>x2-4x-x2+8=0
=>8-4x=0
=>4x=8
=>x=2
b) ( 3x + 2 )( x - 1 ) - 3( x + 1 )( x - 2 ) = 4
=>3x2-x-2-3x2+3x+6=4
=>2x+4=4
=>2x=0
=>x=0
c) x( x + 5 )( x - 5 ) - ( x + 2 )( x^2 - 2x + 4 ) = 17
=>x(x2-25)-(x3+8)=17
=>x3-25x-x3-8=17
=>-25x-8=17
=>-25x=25
=>x=-1
a/ (x + 3)(x - 2) + 3x = 4(x + 3/4)
=> x2 + x - 6 + 3x = 4x + 3
=> x2 = 9 => x = 3 hoặc x = -3
Vậy x = 3 , x = -3
b/ (x2 - 5)(x + 2) + 5x = 2x2 + 17
=> x3 + 2x2 - 5x - 10 + 5x - 2x2 - 17 = 0
=> x3 = 27 => x3 = 33 => x = 3
Vậy x = 3
\(x\left(x-5\right)\left(x+5\right)-\left(x-2\right)\left(x^2+2x+4\right)=-17\)
\(\Leftrightarrow x^3-25x-x^3+8=-17\)
\(\Leftrightarrow-25x=-25\)
hay x=1