1) (3x-2)(9x^2+6x+4)-(2x-5)(2x+5)=(3x-1)^3-(2x+3)^2+9x(3x-1)
Tìm x
Những câu hỏi liên quan
tìm x
1) (3x-2)(9x^2+6x+4)-(2x-5)(2x+5)=(3x-1)^3-(2x+3)^2+9x(3x-1)
2) (2x+1)^3-(3x+2)^2=(2x-5)(4x^2+10x+25)+6x(2x+1)-9x^2
(3x-2)(2x-4)=1-12x²
. Tìm x, biết:
a) 6x.(x – 5) + 3x.(7 – 2x) = 18 b) 2x.(3x + 1) + (4 – 2x).3x = 7 c) 0,5x.(0,4 – 4x) + (2x + 5).x = -6,5 | d) (x + 3)(x + 2) – (x - 2)(x + 5) = 6 e) 3(2x - 1)(3x - 1) – (2x - 3)(9x - 1) = 0 |
a) Ta có: \(6x\left(x-5\right)+3x\left(7-2x\right)=18\)
\(\Leftrightarrow6x^2-30x+21x-6x^2=18\)
\(\Leftrightarrow-9x=18\)
hay x=-2
Vậy: S={-2}
b) Ta có: \(2x\left(3x+1\right)+\left(4-2x\right)\cdot3x=7\)
\(\Leftrightarrow6x^2+2x+12x-6x^2=7\)
\(\Leftrightarrow14x=7\)
hay \(x=\dfrac{1}{2}\)
Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)
c) Ta có: \(0.5x\left(0.4-4x\right)+\left(2x+5\right)\cdot x=-6.5\)
\(\Leftrightarrow0.2x-2x^2+2x^2+5x=-6.5\)
\(\Leftrightarrow5.2x=-6.5\)
hay \(x=-\dfrac{5}{4}\)
Vậy: \(S=\left\{-\dfrac{5}{4}\right\}\)
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d) Ta có: \(\left(x+3\right)\left(x+2\right)-\left(x-2\right)\left(x+5\right)=6\)
\(\Leftrightarrow x^2+5x+6-\left(x^2+3x-10\right)=6\)
\(\Leftrightarrow x^2+5x+6-x^2-3x+10=6\)
\(\Leftrightarrow2x+16=6\)
\(\Leftrightarrow2x=-10\)
hay x=-5
Vậy: S={-5}
e) Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)
\(\Leftrightarrow3\left(6x^2-5x+1\right)-\left(18x^2-29x+3\right)=0\)
\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3=0\)
\(\Leftrightarrow14x=0\)
hay x=0
Vậy: S={0}
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tìm x,y
a) ( 3x-2)^3 - (3x - 2) ( 9x^2 + 6x +4) = 6( 3x+5)(5 - 3x )
b) ( 2x - 1)( 4x^2 - 4x +1) - (2x+ 1)^3+ 3(2x+5)(2x - 5)= -5
tìm x;y: 1) 9x^2-6x=9
2) (2x+3)^2 + 2(2x+3)(x-2)+(2-x)^2=4
3)(x+3)(3-x)=5
4)(3x+2)(9x^2-3x+1)=2
\(\left(2x+3\right)^2+2\left(2x+3\right)\left(x-2\right)+\left(2-x\right)^2=4\)
\(\left(2x+3\right)^2+2\left(2x+3\right)\left(x-2\right)+\left(x-2\right)^2=4\)
\(\left(2x+3+x-2\right)^2=\left(\pm2\right)^2\)
\(\left(3x+1\right)^2=\left(\pm2\right)^2\)
\(\left[\begin{array}{nghiempt}3x+1=2\\3x+1=-2\end{array}\right.\)
\(\left[\begin{array}{nghiempt}3x=2-1\\3x=-2-1\end{array}\right.\)
\(\left[\begin{array}{nghiempt}3x=1\\3x=-3\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=\frac{1}{3}\\x=-1\end{array}\right.\)
***
\(\left(x+3\right)\left(3-x\right)=5\)
\(3^2-x^2=5\)
\(x^2=9-5\)
\(x^2=4\)
\(x^2=\left(\pm2\right)^2\)
\(x=\pm2\)
***
\(\left(3x+1\right)\left(9x^2-3x+1\right)=2\)
\(27x^3+3=2\)
\(27x^3=2-3\)
\(\left(3x\right)^3=-1\)
\(3x=-1\)
\(x=-\frac{1}{3}\)
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Tìm x
a.(x+2).(x+3)-(x-2).(x+5) = 0
b. (2x+3).(x-4)+(x-5)(x+2) = (3x-5)(x-4)
c. (3x+2)(2x+9)-(x+2)(6x+1) = x+1-(x-6)
d. 3( 2x-1).(3x-1)-(2x-3).(9x-1)=0
(x+2)(x+3)-(x-2)(x+5)=0
=> x2+5x+6-x2-3x+10=0
=>2x+16=0
=>2x=-16
=>x=-8
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giải phương trình 1)dfrac{1-6x}{x-2}+dfrac{9x+4}{x+2}dfrac{xleft(3x-2right)+1}{x^2-4}2) dfrac{3x+2}{3x-2}-dfrac{6}{2+3x}dfrac{9x^2}{9x^2-4}3) dfrac{x+5}{3x-6}-dfrac{1}{2}dfrac{2x-3}{2x-4}4) dfrac{x-1}{x}+dfrac{1}{x+1}dfrac{2x-1}{2x^2+2}5) dfrac{2}{x+1}+dfrac{3x+1}{x+1}dfrac{1}{left(x+1right)left(x-2right)}
Đọc tiếp
giải phương trình 1)\(\dfrac{1-6x}{x-2}+\dfrac{9x+4}{x+2}=\dfrac{x\left(3x-2\right)+1}{x^2-4}\)2) \(\dfrac{3x+2}{3x-2}-\dfrac{6}{2+3x}=\dfrac{9x^2}{9x^2-4}\)3) \(\dfrac{x+5}{3x-6}-\dfrac{1}{2}=\dfrac{2x-3}{2x-4}\)4) \(\dfrac{x-1}{x}+\dfrac{1}{x+1}=\dfrac{2x-1}{2x^2+2}\)5) \(\dfrac{2}{x+1}+\dfrac{3x+1}{x+1}=\dfrac{1}{\left(x+1\right)\left(x-2\right)}\)
giúp mình với ạ câu nào cũng được
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Bài 1: giải các pt sau:
1,(x-1)^2-(x+1)^2=2(x+3)
2,(2x-1)^2-(2x+1)^2=4(x-3)
3,(2x+3)^2-(2x+3).(2x-4)=-(x-2)^2
4,8x^3-(x+1)^3=3x-3
5,(3x-2).(9x^2+6x+4)-(3x+1).(9x^2-3x+1)=(2x+1).(2x-1)-4x(x-3)
\(\left(x-1\right)^2-\left(x+1\right)^2=2\left(x+3\right)\)
\(\Leftrightarrow\left(x-1+x+1\right)\left(x-1-x-1\right)=2\left(x+3\right)\)
\(\Leftrightarrow2x\left(-2\right)=2\left(x+3\right)\)
\(\Leftrightarrow-4x=2x+6\)
\(\Leftrightarrow-6x=6\)
\(\Leftrightarrow x=-1\)
2) \(\left(2x-1\right)^2-\left(2x+1\right)^2=4\left(x-3\right)\)
\(\Leftrightarrow\left(2x-1+2x+1\right)\left(2x-1-2x-1\right)-4\left(x-3\right)=0\)
\(\Leftrightarrow4x\left(-2\right)-4x+12=0\)
\(\Leftrightarrow-12x=-12\)
\(\Leftrightarrow x=1\)
3)\(\left(2x+3\right)^2-\left(2x+3\right)\left(2x-4\right)+\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(2x+3\right)\left(2x+3-2x+4\right)+\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow7\left(2x+3\right)+x^2-4x+4=0\)
\(\Leftrightarrow x^2+10x+25=0\)
\(\Leftrightarrow\left(x+5\right)^2=0\)
\(\Leftrightarrow x=-5\)
4) \(8x^3-\left(x+1\right)^3=3x-3\)
\(\Leftrightarrow8x^3-\left(x^3+3x+3x^2+1\right)-3x+3=0\)
\(\Leftrightarrow7x^3-3x^2-6x+2=0\)
\(\Leftrightarrow\left(x-1\right)\left(7x^2+4x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{-2+3\sqrt{2}}{7}\\x=\frac{-2-3\sqrt{2}}{7}\end{matrix}\right.\)
5)\(\left(3x-2\right)\left(9x^2+6x+4\right)-\left(3x-1\right)\left(9x^2-3x+1\right)=x-4\)
\(\Leftrightarrow\left(3x\right)^3-2^3-\left(\left(3x\right)^3-1^3\right)=x-4\)
\(\Leftrightarrow27x^3-8-\left(27x^3-1\right)=x-4\)
\(\Leftrightarrow-7=x-4\)
\(\Leftrightarrow x=-3\)
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Thu gọn biểu thức
(6x+1)(2x-5)
(2x+5)2-2x(2x+8)
(3x-5)(2x-1)-(2x+3)(3x+7)+30x
(X-1)2-(x+1)(x-1)
(3x+2)(9x2-6x+4)-(3+x)(x-3)
(6x+1)(2x-5)=12x2-30x+2x-5=12x2-28x-5
(2x+5)2-2x(2x+8)=4x2+20x+25-4x2-16x=4x+25
(3x-5)(2x-1)-(2x+3)(3x+7)+30x=6x2-3x-10x+5=6x2-13x+5
(X-1)2-(x+1)(x-1)=x2-2x+1-x2+1=-2x+2
(3x+2)(9x2-6x+4)-(3+x)(x-3)=27x3+8+9-x2=27x3-x2+17
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tìm x
a) (x+2)(x+3)-(x-2)(x+5)=6
b) (3x+2)(2x+9)-(x+2)(6x+1)=(x+1)-(x-6)
c) 3(2x-1)(3x-1)-(2x-3)(9x-1)=0
a: Ta có: \(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=6\)
\(\Leftrightarrow x^2+5x+6-x^2-3x+10=6\)
\(\Leftrightarrow2x=-10\)
hay x=-5
b: Ta có: \(\left(3x+2\right)\left(2x+9\right)-\left(x+2\right)\left(6x+1\right)=\left(x+1\right)-\left(x-6\right)\)
\(\Leftrightarrow6x^2+27x+4x+18-6x^2-x-12x-2=x+1-x+6\)
\(\Leftrightarrow18x+16=7\)
hay \(x=-\dfrac{1}{2}\)
c: Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)
\(\Leftrightarrow3\left(6x^2-2x-3x+1\right)-\left(18x^2-2x-27x+3\right)=0\)
\(\Leftrightarrow18x^2-15x+3-18x^2+27x-3=0\)
hay x=0
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