Tìm GTLN
D=4-\(|\)x+1\(|\)-\(|\)4x-3\(|\)-\(|\)x-1\(|\)
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1: Ta có: \(\left(x+3\right)^2-\left(x+2\right)\left(x-2\right)=4x+17\)
\(\Leftrightarrow x^2+6x+9-x^2+4-4x=17\)
\(\Leftrightarrow x=2\)
3: Ta có: \(\left(2x+3\right)\left(x-1\right)+\left(2x-3\right)\left(1-x\right)=0\)
\(\Leftrightarrow2x^2-2x+3x-3+2x-2x^2-3+3x=0\)
\(\Leftrightarrow6x=6\)
hay x=1
2:
=>x^3-1-2x^3-4x^6+4x^6+4x=6
=>-x^3+4x-7=0
=>x=-2,59
4: =>8x-24x^2+2-6x+24x^2-60x-4x+10=-50
=>-62x+12=-50
=>x=1
n: ĐKXĐ: x<>0
\(\left(x+\dfrac{1}{x}\right)^2-3\left(x+\dfrac{1}{x}\right)+2=0\)
=>\(\left(x+\dfrac{1}{x}\right)^2-2\left(x+\dfrac{1}{x}\right)-\left(x+\dfrac{1}{x}\right)+2=0\)
=>\(\left(x+\dfrac{1}{x}-2\right)\left(x+\dfrac{1}{x}-1\right)=0\)
=>\(\dfrac{x^2+1-2x}{x}\cdot\dfrac{x^2+1-x}{x}=0\)
=>\(\left(x^2-2x+1\right)\left(x^2-x+1\right)=0\)
=>\(\left(x-1\right)^2\left[\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right]=0\)
=>\(\left(x-1\right)^2=0\)
=>x-1=0
=>x=1
p: \(x^4-4x^3+6x^2-4x+1=0\)
=>\(x^4-x^3-3x^3+3x^2+3x^2-3x-x+1=0\)
=>\(x^3\left(x-1\right)-3x^2\left(x-1\right)+3x\left(x-1\right)-\left(x-1\right)=0\)
=>\(\left(x-1\right)\left(x^3-3x^2+3x-1\right)=0\)
=>\(\left(x-1\right)^4=0\)
=>x-1=0
=>x=1
a: ta có: \(\left(8x+2\right)\left(1-3x\right)+\left(6x-1\right)\left(4x-10\right)=-50\)
\(\Leftrightarrow8x-24x^2+2-6x+24x^2-60x-4x+40=-50\)
\(\Leftrightarrow-62x=-92\)
hay \(x=\dfrac{46}{31}\)
b: ta có: \(\left(1-4x\right)\left(x-1\right)+4\left(3x+2\right)\left(x+3\right)=38\)
\(\Leftrightarrow x-1-4x^2+4x+4\left(3x^2+9x+2x+6\right)=38\)
\(\Leftrightarrow-4x^2+5x-1+12x^2+44x+24-38=0\)
\(\Leftrightarrow8x^2+49x-15=0\)
\(\text{Δ}=49^2-4\cdot8\cdot\left(-15\right)=2881\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{-49-\sqrt{2881}}{16}\\x_2=\dfrac{-49+\sqrt{2881}}{16}\end{matrix}\right.\)
`2//(5x-8)-3(4x-5)=4(3x-4)`
`<=>5x-8-12x+15=12x-16`
`<=>-19x=-23`
`<=>x=23/19` Vậy `x=23/19`
`3//2(x^3-1)-2x^2(x+2x^4)+(4x^5+4)x=6`
`<=>2x^3-2-2x^3-4x^6+4x^6+4x=6`
`<=>4x=8`
`<=>x=2` Vậy `x=2`
a: Ta có: \(\left(8x^2-4x\right):\left(-4x\right)-\left(x+2\right)=8\)
\(\Leftrightarrow-2x+1-x-2=8\)
\(\Leftrightarrow-3x=9\)
hay x=-3
b: Ta có: \(\left(2x^4-3x^3+x^2\right):\left(-\dfrac{1}{2}x^2\right)+4\left(x-1\right)^2=0\)
\(\Leftrightarrow-4x^2+6x-2+4x^2-8x+4=0\)
\(\Leftrightarrow-2x=-2\)
hay x=1