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Xin lỗi bạn,mk ms học đến phân tích đa thức thành nhân tử nhóm nhiều hạng tử,còn phần này mk ms học còn yếu lắm.
1. \(-10x^2+11x+6\)
\(=-10x^2+15x-4x+6\)
\(=-5x\left(2x-3\right)-2\left(2x-3\right)\)
\(=\left(-5x-2\right)\left(2x-3\right)\)
2.\(10x^2-4x-6\)
\(=2\left(5x^2-2x-3\right)\)
\(=2\left(5x^2+3x-5x-3\right)\)
\(=2\left[x\left(5x+3\right)-\left(5x+3\right)\right]\)
\(=2\left(x-1\right)\left(5x+3\right)\)
3. \(10x^2+7x-6\)
\(=10x^2+12x-5x-6\)
\(=2x\left(5x+6\right)-\left(5x+6\right)\)
\(=\left(2x-1\right)\left(5x+6\right)\)
4. \(10x^2-14x-12\)
\(=2\left(5x^2-7x-6\right)\)
\(=2\left(5x^2+3x-10x-6\right)\)
\(=2\left[x\left(5x+3\right)-2\left(5x+3\right)\right]\)
\(=2\left(x-2\right)\left(5x+3\right)\)
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a) \(10x^2-29x+10\)
\(=10x^2-4x-25x+10\)
\(=2x\left(5x-2\right)-5\left(5x-2\right)\)
\(=\left(5x-2\right)\left(2x-5\right)\)
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a: \(\Leftrightarrow\left(x+12-3x\right)\left(x+12+3x\right)=0\)
=>(-2x+12)(4x+12)=0
=>x=-3 hoặc x=6
b: \(\Leftrightarrow20x^3-15x^2+45x-45=0\)
=>\(x\simeq0.93\)
d: =>-4x+28+11x=-x+3x+15
=>7x+28=2x+15
=>5x=-13
=>x=-13/5
e: \(\Leftrightarrow4x^3-12x+x=4x^3-3x+5\)
=>-9x=-3x+5
=>-6x=5
=>x=-5/6
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a. ĐKXĐ: \(x\ne1\)
\(\dfrac{11x-4}{x-1}+\dfrac{10x+4}{2-2x}=\dfrac{2\cdot\left(11x-4\right)}{2\cdot\left(x-1\right)}-\dfrac{10x+4}{2x-2}\)
\(=\dfrac{22x-8}{2\left(x-1\right)}-\dfrac{10x+4}{2\left(x-1\right)}\)\(=\dfrac{22x-8-10x-4}{2\left(x-1\right)}\)
\(=\dfrac{12x-12}{2\left(x-1\right)}\)\(=\dfrac{12\left(x-1\right)}{2\left(x-1\right)}=6\)
b. ĐKXĐ: \(x\ne-2;x\ne\dfrac{1}{2}\)
\(\dfrac{1}{x+2}+\dfrac{5}{2x^2+3x-2}=\dfrac{2x-1}{\left(2x-1\right)\left(x+2\right)}+\dfrac{5}{\left(2x-1\right)\left(x+2\right)}\)
\(=\dfrac{2x-1+5}{\left(2x-1\right)\left(x+2\right)}=\dfrac{2x+4}{\left(2x-1\right)\left(x+2\right)}\)
\(=\dfrac{2\left(x+2\right)}{\left(2x-1\right)\left(x+2\right)}=\dfrac{2}{2x-1}\)
\(\text{#}Toru\)
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1.3x^2-11x+6=3x^2-9x-2x+6=3x(x-3)-2(x-3)=(3x-2)(x-3)
2.8x^2+10x-3=8x^2-2x+12x-3=2x(4x-1)+3(4x-1)=(2x+3)(4x-1)
1. 10x2-9x-1
= 10x2-10x+x-1
= 10x(x-1)+(x-1)
= (x-1)(10x+1)
2. 11x2-10x-1
= 11x2-11x+x-1
= 11x(x-1)+(x-1)
= (x-1)(11x+1)