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\(a,\Leftrightarrow x^2+6x+9-x^2+3x+10=1\\ \Leftrightarrow9x=-18\Leftrightarrow x=-2\\ b,\Leftrightarrow4x^2-4x+1-4x^2+17x+15=3\\ \Leftrightarrow13x=-13\Leftrightarrow x=-1\\ c,\Leftrightarrow3x\left(x-2\right)+4\left(x-2\right)=0\\ \Leftrightarrow\left(3x+4\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=2\end{matrix}\right.\\ d,\Leftrightarrow2x\left(3x+5\right)-6\left(3x+5\right)=0\\ \Leftrightarrow\left(x-3\right)\left(3x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{3}\end{matrix}\right.\)
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) Ta có: \(5x\left(x+1\right)-5\left(x+1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left[5x-5\left(x-2\right)\right]=0\)
\(\Leftrightarrow\left(x+1\right)\left(5x-5x+10\right)=0\)
\(\Leftrightarrow10\left(x+1\right)=0\)
mà \(10\ne0\)
nên x+1=0
hay x=-1
Vậy: x=-1
b) Ta có: \(\left(4x+1\right)\left(x-2\right)-\left(2x-3\right)=4\)
\(\Leftrightarrow4x^2-8x+x-2-2x+3-4=0\)
\(\Leftrightarrow4x^2-9x-3=0\)
\(\Leftrightarrow\left(2x\right)^2-2\cdot2x\cdot\frac{9}{4}+\frac{81}{16}-\frac{129}{16}=0\)
\(\Leftrightarrow\left(2x-\frac{9}{4}\right)^2=\frac{129}{16}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{9}{4}=\frac{\sqrt{129}}{4}\\2x-\frac{9}{4}=-\frac{\sqrt{129}}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\frac{9+\sqrt{129}}{4}\\2x=\frac{9-\sqrt{129}}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{9+\sqrt{129}}{8}\\x=\frac{9-\sqrt{129}}{8}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{9+\sqrt{129}}{8};\frac{9-\sqrt{129}}{8}\right\}\)
c) Ta có: \(2x^3-18x=0\)
\(\Leftrightarrow2x\left(x^2-9\right)=0\)
\(\Leftrightarrow2x\left(x-3\right)\left(x+3\right)=0\)
mà \(2\ne0\)
nên \(\left[{}\begin{matrix}x=0\\x+3=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=3\end{matrix}\right.\)
Vậy: \(x\in\left\{0;-3;3\right\}\)
d) Ta có: \(\left(3x-2\right)\left(2x+1\right)-6x\left(x+2\right)=11\)
\(\Leftrightarrow6x^2+3x-4x-2-6x^2-12x=11\)
\(\Leftrightarrow-13x-2=11\)
\(\Leftrightarrow-13x=13\)
hay x=-1
Vậy: x=-1
e) Ta có: \(\left(x-1\right)^3-\left(x+2\right)\left(x^2-2x+4\right)=3\left(1-x^2\right)\)
\(\Leftrightarrow x^3-3x^2+3x-1-\left(x^3+8\right)=3-3x^2\)
\(\Leftrightarrow x^3-3x^2+3x-1-x^3-8-3+3x^2=0\)
\(\Leftrightarrow3x-12=0\)
\(\Leftrightarrow3x=12\)
hay x=4
Vậy: x=4
f) Ta có: \(6x^2-\left(2x+5\right)\left(3x-2\right)=-1\)
\(\Leftrightarrow6x^2-\left(6x^2-4x+15x-10\right)+1=0\)
\(\Leftrightarrow6x^2-6x^2+4x-15x+10+1=0\)
\(\Leftrightarrow-11x+11=0\)
\(\Leftrightarrow-11x=-11\)
hay x=1
Vậy: x=1
1: \(\dfrac{2x^3+11x^2+18x-3}{2x+3}\)
\(=\dfrac{2x^3+3x^2+8x^2+12x+6x+9-12}{2x+3}\)
\(=x^2+4x+3-\dfrac{12}{2x+3}\)
+) (5x-1). (2x+3)-3. (3x-1)=0
10x^2+15x-2x-3 - 9x+3=0
10x^2 +8x=0
2x(5x+4)=0
=> x=0 hoặc x= -4/5
+) x^3 (2x-3)-x^2 (4x^2-6x+2)=0
2x^4 -3x^3 -4x^4 + 6x^3 - 2x^2=0
-2x^4 + 3x^3-2x^2=0
x^2(-2x^2+x-2)=0
-2x^2(x-1)^2=0
=> x=0 hoặc x=1
+) x (x-1)-x^2+2x=5
x^2 -x -x^2+2x=5
x=5
+) 8 (x-2)-2 (3x-4)=25
8x - 16-6x+8=25
2x=33
x=33/2
a) 3x(4x-3)-2x(5-6x)=0
\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)
\(\Leftrightarrow24x^2-19x=0\)
\(\Leftrightarrow x\left(24x-19\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\24x-19=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\24x=19\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{19}{24}\end{matrix}\right.\)
Vậy x=0 hoặc x=\(\dfrac{19}{24}\)
a. 5x.(12x+7)-3x.(20x-5)=-150
x=-3
b. ( 2x-1).(3-x)+(x+4).(2x-5)=20
x=43/10
c. 9x2-1+(3x-1)2=0
x=1/3
d. 3x.(x-2)-(3x+2).(x-1)=7
x=-5/2
e. (2x-1)2-(2x+5).(2x-5)=20
x=3/2
f. 4x2-5=4
x=3/2
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giai phuong trinh
1, (x-2)(x-1)(x-8)(x-4)=4x^2
2, (x^2+5x+6)(x^2+20x+96)=4x^2
3, 3(x^2+2x-1)^2-2(x^2+3x-1)^2+5x^2=0
x2-4x+5=0
=>(x-2)2+1=0
=>(x-2)2 =-1
=> pt vô nghiệm
(x2+5x)(x3+3x2-18x)=0
=>\(\int^{x^2+5x=0}_{x^3+3x^2-18x=0}=>\int^{\int^{x=0}_{x=-5}}_{x=3;x=0;x=-6}\)