\(\left(3x-1\right)^2-5\left(6x-3\right)\left(2x+1\right)=\left(x-1\right)^2\)
TH
Những câu hỏi liên quan
Chứng minh biểu thức không phụ thuộc x :
1, \(\left(3x-1\right)^2-2\cdot\left(2x-3\right)\cdot\left(2x+3\right)-\left(x-3\right)^2\)
2, \(\left(3x+2\right)^3-\left(3x-2\right)^3-3\cdot\left(6x-1\right)\cdot\left(6x+1\right)\)
3, \(\left(3x-5\right)^2+3\cdot\left(x+1\right)\cdot\left(x-1\right)-\left(4x-3\right)^2+\left(2x+2\right)\cdot\left(2x+1\right)\)
\(\left(3x-1\right)^2-5\left(2x+1\right)^2+\left(6x-3\right)\left(2x+1\right)=\left(x-1\right)^2\)
\(\left(3x-1\right)^2-5\left(2x+1\right)^2+\left(6x-3\right)\left(2x+1\right)=\left(x-1\right)^2\\ \Leftrightarrow\left(9x^2-6x+1\right)-\left(20x^2+20x+5\right)+\left(12x^2-3\right)-\left(x^2-2x+1\right)=0\\ \Leftrightarrow12x-8=0\\ \Leftrightarrow x=\dfrac{2}{3}\)
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BÀI 6 tìm x1,2xleft(x-5right)-left(3x+2x^2right)0 2,xleft(5-2xright)+2xleft(x-1right)133,2x^3left(2x-3right)-x^2left(4x^2-6x+2right)0 4,5xleft(x-1right)-left(x+2right)left(5x-7right)65,6x^2-left(2x-3right)left(3x+2right)1 6,2xleft(1-xright)+59-2x^2
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BÀI 6 tìm x
1,\(2x\left(x-5\right)-\left(3x+2x^2\right)=0\) 2,\(x\left(5-2x\right)+2x\left(x-1\right)=13\)
3,\(2x^3\left(2x-3\right)-x^2\left(4x^2-6x+2\right)=0\) 4,\(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
5,\(6x^2-\left(2x-3\right)\left(3x+2\right)=1\) 6,\(2x\left(1-x\right)+5=9-2x^2\)
1: \(\Leftrightarrow2x^2-10x-3x-2x^2=0\)
=>-13x=0
=>x=0
2: \(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
=>3x=13
=>x=13/3
3: \(\Leftrightarrow4x^4-6x^3-4x^3+6x^3-2x^2=0\)
=>-2x^2=0
=>x=0
4: \(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
=>-8x=6-14=-8
=>x=1
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`1)2x(x-5)-(3x+2x^2)=0`
`<=>2x^2-10x-3x-2x^2=0`
`<=>-13x=0`
`<=>x=0`
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`2)x(5-2x)+2x(x-1)=13`
`<=>5x-2x^2+2x^2-2x=13`
`<=>3x=13<=>x=13/3`
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`3)2x^3(2x-3)-x^2(4x^2-6x+2)=0`
`<=>4x^4-6x^3-4x^4+6x^3-2x^2=0`
`<=>x=0`
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`4)5x(x-1)-(x+2)(5x-7)=0`
`<=>5x^2-5x-5x^2+7x-10x+14=0`
`<=>-8x=-14`
`<=>x=7/4`
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`5)6x^2-(2x-3)(3x+2)=1`
`<=>6x^2-6x^2-4x+9x+6=1`
`<=>5x=-5<=>x=-1`
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`6)2x(1-x)+5=9-2x^2`
`<=>2x-2x^2+5=9-2x^2`
`<=>2x=4<=>x=2`
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Tìm x, biết :
a, \(\left(3x+2\right).\left(6x-2\right)-\left(9x-2\right).\left(2x+1\right)=24\)
b, \(\left(4x+3\right)\left(3x-2\right)-\left(6x-1\right)\left(2x+3\right)=16\)
c, \(\left(5x-2\right)\left(4x+5\right)+\left(10x-7\right)\left(5-2x\right)=12\)
d, \(6x\left(3-4x\right)+8x\left(3x-2\right)=16\)
giải pt:
a,\(\left(13-4x\right)\sqrt{2x-3}+\left(4x-3\right)\sqrt{5-2x}=2+8\sqrt{-4x^2+16x-15}\)
b,\(\left(9x-2\right)\sqrt{3x-1}+\left(10-9x\right)\sqrt{3-3x}-4\sqrt{-9x^2+12x-3}=4\)
c, \(\left(6x-5\right)\sqrt{x+1}-\left(6x+2\right)\sqrt{x-1}+4\sqrt{x^2-1}=4x-3\)
giải các phương trình sau
\(\left(3x-1\right)\left(2x+7\right)-\left(x+1\right)\left(6x-5\right)=\)16
\(\left(2x+3\right)^2-2\left(2x+3\right)\left(2x-5\right)+\left(2x-5\right)^2=x^2+6x+64\)
\(\left(x^4+2x^3+10x-25\right):\left(x^2+5\right)=3\)
giải pt :
a,\(\left(6x-5\right)\sqrt{x+1}-\left(6x+2\right)\sqrt{x-1}+4\sqrt{x^2-1}=4x-3\)
b, \(\left(9x-2\right)\sqrt{3x-1}+\left(10-9x\right)\sqrt{3-3x}-4\sqrt{-9x^2+12x-3}=4\)
c, \(\left(13-4x\right)\sqrt{2x-3}+\left(4x-3\right)\sqrt{5-2x}=2+8\sqrt{-4x^2+16x-15}\)
Bài 1 : tìm các giá trị của x biết :
a) \(\left(3x-5\right)\left(2x-1\right)-\left(x+2\right)\left(6x-1\right)=0\)
b) \(\left(3x-2\right)\left(3x+2\right)-\left(3x-1\right)^2=-5\)
c) \(x^2=-6x-8\)
d) \(\frac{\left(x+1\right)^2}{3}-\frac{\left(x-2\right)^2}{3}=\frac{2x+1}{2}-\frac{\left(x-3\right)^2}{6}\)
a) \(\left(3x-5\right)\left(2x-1\right)-\left(x+2\right)\left(6x-1\right)=0\)
⇔ \(6x^2-13x+5-6x^2-11x+2=0\)
⇔ \(24x=7\)⇔\(x=\frac{7}{24}\)
b) \(\left(3x-2\right)\left(3x+2\right)-\left(3x-1\right)^2=-5\)
⇔ \(9x^2-4-9x^2+6x-1=5\)
⇔ \(6x=10\)⇔ \(x=\frac{5}{3}\)
c) \(x^2=-6x-8\)⇔\(x^2+6x+8=0\)⇔\(\left(x+2\right)\left(x+4\right)=0\)
⇔\(\left[{}\begin{matrix}x=-2\\x=-4\end{matrix}\right.\)
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Bài 1 : tìm các giá trị của x biết :
a) \(\left(3x-5\right)\left(2x-1\right)-\left(x+2\right)\left(6x-1\right)=0\)
b) \(\left(3x-2\right)\left(3x+2\right)-\left(3x-1\right)^2=-5\)
c) \(x^2=-6x-8\)
d) \(\frac{\left(x+1\right)^2}{3}-\frac{\left(x-2\right)^2}{3}=\frac{2x+1}{2}-\frac{\left(x-3\right)^2}{6}\)
a, (3x - 5)(2x - 1) - (x + 2)(6x - 1) = 0
=> 6x^2 - 3x - 10x + 5 - (6x^2 - x + 12x - 2) = 0
=> 6x^2 - 13x + 5 - 6x^2 - 11x + 2 = 0
=> -24x + 7 = 0
=> - 24x = -7
=> x = 7/24
b, (3x - 2)(3x + 2) - (3x - 1)^2 = -5
=> 9x^2 - 4 - 9x^2 + 6x - 1 = -5
=> 6x - 5 = -5
=> 6x = 0
=> x = 0
c, x^2 = -6x - 8
=> x^2 + 6x + 8 = 0
=> x^2 + 2.x.3 + 9 - 1 = 0
=> (x + 3)^2 = 1
=> x + 3 = 1 hoặc x + 3 = -1
=> x = -2 hoặc x = -4
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