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`a)2x^2+3(x-1)(x+1)=5x(x+1)`
`<=>2x^2+3x^2-3=5x^2+5x`
`<=>5x=-3`
`<=>x=-3/5`
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`b)(x-3)^3+3-x=0` nhỉ?
`<=>(x-3)^3-(x-3)=0`
`<=>(x-3)(x^2-1)=0`
`<=>[(x=3),(x^2=1<=>x=+-1):}`
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`c)5x(x-2000)-x+2000=0`
`<=>5x(x-2000)-(x-2000)=0`
`<=>(x-2000)(5x-1)=0`
`<=>[(x=2000),(x=1/5):}`
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`d)3(2x-3)+2(2-x)=-3`
`<=>6x-9+4-2x=-3`
`<=>4x=2`
`<=>x=1/2`
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`e)x+6x^2=0`
`<=>x(1+6x)=0`
`<=>[(x=0),(x=-1/6):}`
= x3 + 33 -x(x2 -1) -27 =0 ( tổng các lập phuong)
x =0
CX100%
\(\Leftrightarrow x^2-6x+9-x^2+4=1\)
=>-6x=-12
hay x=2
\(A=3\left(x-3\right)\left(x+7\right)+\left(x+4\right)^2+48\)
\(A=3\left(x^2-4x-21\right)+\left(x^2+8x+16\right)+48\)
\(A=\left(3x^2+x^2\right)-\left(12x-8x\right)-\left(21-16-48\right)\)
\(A=4x^2-4x+43\)
\(A=\left(4x^2-4x+1\right)+42\)
\(A=\left(2x+1\right)^2+42\)
Thay \(x=\frac{1}{2}\) vao A ta duoc:
\(A=\left(2\cdot\frac{1}{2}+1\right)^2+42=46\)
\(A=3\left(x-3\right)\left(x-7\right)+\left(x+4\right)^2+48\)
\(=3x^2-13x+63+x^2+8x+16+48\)
\(=4x^2-5x+127\)
\(4\cdot0,25-5\cdot0,5+127=1-1+127=127\)
1)
\((x+2)(x+3)(x+4)(x+5)-24\\=[(x+2)(x+5)]\cdot[(x+3)(x+4)]-24\\=(x^2+7x+10)(x^2+7x+12)-24\)
Đặt \(x^2+7x+10=y\), khi đó biểu thức trở thành:
\(y(y+2)-24\\=y^2+2y-24\\=y^2+2y+1-25\\=(y+1)^2-5^2\\=(y+1-5)(y+1+5)\\=(y-4)(y+6)\\=(x^2+7x+10-4)(x^2+7x+10+6)\\=(x^2+7x+6)(x^2+7x+16)\)
2) Bạn xem lại đề!
1
(x2-8)2+36
=x4-16x2+64+36
=x4+20x2+100-36x2
=(x2+10)2-(6x)2
HĐT số 3
Câu 1:
\(=x^4-16x^2+64+36\)
\(=x^4-16x^2+100\)
\(=x^4+20x^2+100-36x^2\)
\(=\left(x^2+10\right)^2-\left(6x\right)^2\)
\(=\left(x^2-6x+10\right)\left(x^2+6x+10\right)\)
Câu 2: \(=x^4+2x^2+1-x^2\)
\(=\left(x^2+1\right)^2-x^2\)
\(=\left(x^2+x+1\right)\left(x^2-x+1\right)\)
\(\dfrac{1+x}{1-x}+3=\dfrac{x-3}{x-1}\)
\(ĐK:x\ne1\)
\(\Leftrightarrow\dfrac{1+x}{1-x}+3=\dfrac{3-x}{1-x}\)
\(\Leftrightarrow\dfrac{\left(1+x\right)+3\left(1-x\right)}{1-x}=\dfrac{3-x}{1-x}\)
\(\Leftrightarrow\left(1+x\right)+3\left(1-x\right)=3-x\)
\(\Leftrightarrow1+x+3-3x=3-x\)
\(\Leftrightarrow-x=-1\)
\(\Leftrightarrow x=1\left(ktm\right)\)
Vậy pt vô nghiệm
\(\dfrac{1+x}{1-x}+3=\dfrac{x-3}{x-1}\) đề như thế này phải ko?