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(5/7x-1/4)(-3/4x+1/2)=0
\(\Rightarrow\left[{}\begin{matrix}\dfrac{5}{7}x-\dfrac{1}{4}=0\\-\dfrac{3}{4}x+\dfrac{1}{2}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\dfrac{5}{7}x=\dfrac{1}{4}\\-\dfrac{3}{4}x=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}:\dfrac{5}{7}\\x=\dfrac{1}{2}:\dfrac{3}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}.\dfrac{7}{5}\\x=\dfrac{1}{2}.\dfrac{4}{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{20}\\x=\dfrac{4}{6}=\dfrac{2}{3}\end{matrix}\right.\)
Vậy ....
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a: 3-2|4x-5|=2/6
=>2|4x-5|=3-1/3=8/3
=>|4x-5|=4/3
=>4x-5=4/3 hoặc 4x-5=-4/3
=>4x=19/3 hoặc 4x=11/3
=>x=19/12 hoặc x=11/12
c: (7-3x)(2x+1)=0
=>2x+1=0 hoặc -3x+7=0
=>x=-1/2 hoặc x=-7/3
d: 2x(5-3x)>0
=>x(3x-5)<0
=>0<x<5/3
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1. \(A=x^{15}+3x^{14}+5=x^{14}\left(x+3\right)+5\)
Thay \(x+3=0\)vào đa thức ta được:\(A=x^{14}.0+5=5\)
2. \(B=\left(x^{2007}+3x^{2006}+1\right)^{2007}=\left[x^{2006}\left(x+3\right)+1\right]^{2007}\)
Thay \(x=-3\)vào đa thức ta được: \(B=\left[x^{2006}\left(-3+3\right)+1\right]^{2017}=\left(x^{2006}.0+1\right)^{2017}=1^{2017}=1\)
3. \(C=21x^4+12x^3-3x^2+24x+15=3x\left(7x^3+4x^2-x+8\right)+15\)
Thay \(7x^3+4x^2-x+8=0\)vào đa thức ta được: \(C=3x.0+15=15\)
4. \(D=-16x^5-28x^4+16x^3-20x^2+32x+2007\)
\(=4x\left(-4x^4-7x^3+4x^2-5x+8\right)+2007\)
Thay \(-4x^4-7x^3+4x^2-5x+8=0\)vào đa thức ta được: \(D=4x.0+2007=2007\)
1. \(A=x^{15}+3x^{14}+5\)
\(A=x^{14}\left(x+3\right)+5\)
\(A=x^{14}+5\)
2. \(B=\left(x^{2007}+3x^{2006}+1\right)^{2007}\)
\(B=\left[x^{2006}\left(x+3\right)+1\right]^{2007}\)
\(B=\left[x^{2006}.\left(-3+3\right)+1\right]^{2007}\)
\(B=1^{2007}=1\)
3. \(C=21x^4+12x^3-3x^2+24x+15\)
\(C=3x\left(7x^2+4x^2-x+8+5\right)\)
\(C=3x\left(0+5\right)\)
\(C=15x\)
4. \(D=-16x^5-28x^4+16x^3-20x^2+32+2007\)
\(D=4x\left(-4x^4-7x^3+4x^2-5x+8\right)+2007\)
\(D=4x.0+2007\)
\(D=2007\)
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Bài 1:
a) \(-5\left(x^2-3x+1\right)+x\left(1+5x\right)=x-2\)
\(\Rightarrow-5x^2+15x-5+x+5x^2=x-2\)
\(\Rightarrow16x-5=x-2\)
\(\Rightarrow16x-x=5-2\)
\(\Rightarrow15x=3\)
\(\Rightarrow x=\dfrac{15}{3}=5\)
b) \(12x^2-4x\left(3x+5\right)=10x-17\)
\(\Rightarrow12x^2-12x^2-20x=10x-17\)
\(\Rightarrow-20x=10x-17\)
\(\Rightarrow-20x-10x=-17\)
\(\Rightarrow-30x=-17\)
\(\Rightarrow x=\dfrac{-30}{-17}=\dfrac{30}{17}\)
c) \(-4x\left(x-5\right)+7x\left(x-4\right)-3x^2=12\)
\(\Rightarrow-4x^2+20x+7x^2-28x-3x^2=12\)
\(\Rightarrow-8x=12\)
\(\Rightarrow x=\dfrac{12}{-8}=-\dfrac{4}{3}\)
Bài 2:
a) \(\left(x+5\right)\left(x-7\right)-7x\left(x-3\right)\)
\(=x^2-7x+5x-35-7x^2+21x\)
\(=-6x^2+19x-35\)
b) \(x\left(x^2-x-2\right)-\left(x-5\right)\left(x+1\right)\)
\(=x^3-x^2-2x-x^2+x-5x-5\)
\(=x^3-2x^2-6x-5\)
c) \(\left(x-5\right)\left(x-7\right)-\left(x+4\right)\left(x-3\right)\)
\(=x^2-7x-5x+35-x^2-3x+4x-12\)
\(=11x+23\)
d) \(\left(x-1\right)\left(x-2\right)-\left(x+5\right)\left(x+2\right)\)
\(=x^2-2x-x+2-x^2+2x+5x+10\)
\(=4x+12\)
\(\left(\frac{5}{7}x-\frac{1}{4}\right)\left(\frac{-3}{4}x+\frac{1}{2}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{5}{7}x-\frac{1}{4}=0\\\frac{-3}{4}x+\frac{1}{2}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{20}\\x=\frac{2}{3}\end{cases}}\)