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<=> 4(x^2 + 2x + 1) + 4x^2 - 4x +1 - 8(x^2 - 1) = 11
<=> 4x^2 + 8x + 4 + 4x^2 - 4x +1 - 8x^2 +8 - 11 = 0
<=> 4x + 2 = 0
<=> x = - 1/2
`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):}`
áp dụng BĐT cô si a ^2+b^2=2ab ta dc
x^2+y^2>=2xy
x^2+1>2.x.1=2x
y^2+1=2.y.1=2y
cộng theo vế 3 ĐBTtrên ta dc x^2+y^2+x^2+1>=2xy+2x+2y
(=)2(x^2+y^2+1>=2(xy+x+y)
(=)x^2+y^2+1=xy+x+y
a,gt <=> (x-1)^3-x(x^2-4)=5
<=>x^3-3x^2+3x-1-x^3+4x-5=0
<=>-3x^2+7x-6=0
<=>x^2-7/3x+2=0
<=>x^2-7/3x+49/36+23/36=0
<=>(x-7/6)^2=-23/36(vl)
=>không tìm được x
b,gt <=> (x-1)^3-(x+3)^3+3(x^2-4)=2
<=>x^3-3x^2+3x-1-x^3-9x^2-27x-27+3x^2-12-2=0
<=>-9x^2-24x-42=0
<=>9x^2+24x+16=-26
<=>(3x+4)^2=-26(vl)
=> điều tương tự câu a
toán lớp 8 ah răng lại hỏi chắc là câu ni anh mi ko biết hầy
Ta có : (x + 1)2 - (x + 2)(x - 2) = 0
<=> (x + 1)2 - (x2 - 22) = 0
<=> x2 + 2x + 1 - x2 + 4 = 0
<=> 2x + 5 = 0
=> 2x = -5
=> x = \(-\frac{5}{2}\)
\(\Leftrightarrow\frac{1}{x^2+5x+6}+\frac{1}{x^2+3x+2}+\frac{1}{x^2+x}=\frac{3}{10}\)
\(\Leftrightarrow\frac{1}{x^2+5x+6}+\frac{1}{x^2+3x+2}+\frac{1}{x^2+x}-\frac{3}{10}=0\)
\(\Leftrightarrow-\frac{3\left(x^2+3x-10\right)}{10x\left(x+3\right)}=0\)
\(\Leftrightarrow3\left(x^2+3x-10\right)=0\)
\(\Leftrightarrow x^2+3x-10=0\)
\(\Leftrightarrow x^2-2x+5x-10=0\)
\(\Leftrightarrow x\left(x-2\right)+5\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow x-2=0\)hoặc\(x+5=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-5\\x=2\end{cases}}\)
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}=\frac{3}{10}\)
\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}=\frac{3}{10}\)
\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+3}=\frac{3}{10}\Leftrightarrow\frac{\left(x+3\right)-x}{x\left(x+3\right)}=\frac{3}{10}\Leftrightarrow\frac{3}{x\left(x+3\right)}=\frac{3}{10}\)
<=>x(x+3)=10 <=> x2+3x=10 <=> x2+3x-10=0
<=>-(x2-3x+10)=0
<=>x2-3x+10=0
<=>x2-2.x.\(\frac{3}{2}\)+ \(\left(\frac{3}{2}\right)^2+\frac{31}{4}\)=0
<=> \(\left(x-\frac{3}{2}\right)^2+\frac{31}{4}\)=0
Vì \(\left(x-\frac{3}{2}\right)^2\ge0\Rightarrow\left(x-\frac{3}{2}\right)^2+\frac{31}{4}\ge\frac{31}{4}>0\) (với mọi x)
=>PT vô nghiệm