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\(A=x^2-4x-x\left(x-4\right)-15\)
\(=x^2-4x-x^2+4x-15=-15\) => đpcm
\(B=5x\left(x^2-x\right)-x^2\left(5x-5\right)-13\)
\(=5x^3-5x^2-5x^3+5x^2-13=-13\) => đpcm
\(C=-3x\left(x-5\right)+3\left(x^2-4x\right)-3x+7\)
\(=-3x^2+15x+3x^2-12x-3x+7=7\) => đpcm
\(D=7\left(x^2-5x+3\right)-x\left(7x-35\right)-14\)
\(=7x^2-35x+21-7x^2+35x-14=7\) => đpcm
\(E=4x\left(x^2-7+2\right)-4\left(x^3-7x+2x-5\right)\)
\(=4x^3-20x-4x^3+20x+20=20\) => đpcm
\(H=x\left(5x-3\right)-x^2\left(x-1\right)+x\left(x^2-6x\right)-10+3x\)
\(=5x^2-3x-x^3+x^2+x^3-6x^2-10x+3x=-10\) => đpcm
a) Ta có: \(\frac{3x-11}{11}-\frac{x}{3}=\frac{3x-5}{7}-\frac{5x-3}{9}\)
\(\Leftrightarrow\frac{63\left(3x-11\right)}{693}-\frac{231x}{693}-\frac{99\left(3x-5\right)}{693}+\frac{77\left(5x-3\right)}{693}=0\)
\(\Leftrightarrow189x-693-231x-297x+495+385x-231=0\)
\(\Leftrightarrow46x-429=0\)
\(\Leftrightarrow46x=429\)
hay \(x=\frac{429}{46}\)
Vậy: \(x=\frac{429}{46}\)
b) Ta có: \(\frac{9x-0,7}{4}-\frac{5x-1,5}{7}=\frac{7x-1,1}{6}-\frac{5\left(0,4-2x\right)}{5}\)
\(\Leftrightarrow\frac{9x-0,7}{4}-\frac{5x-1,5}{7}-\frac{7x-1,1}{6}+\frac{5\left(0,4-2x\right)}{5}=0\)
\(\Leftrightarrow105\left(9x-0,7\right)-60\left(5x-1,5\right)-70\left(7x-1,1\right)+420\left(0,4-2x\right)=0\)
\(\Leftrightarrow945x-\frac{147}{2}-300x+90-490x+77+168-840x=0\)
\(\Leftrightarrow-685x+261.5=0\)
\(\Leftrightarrow-685x=-261.5\)
hay \(x=\frac{523}{1370}\)
Vậy: \(x=\frac{523}{1370}\)
c) Ta có: \(\frac{5\left(x-1\right)+2}{6}-\frac{7x-1}{4}=\frac{2\left(2x-1\right)}{7}-5\)
\(\Leftrightarrow\frac{14\left(5x-3\right)}{84}-\frac{21\left(7x-1\right)}{84}-\frac{24\left(2x-1\right)}{84}+\frac{420}{84}=0\)
\(\Leftrightarrow70x-42-147x+21-48x+24+420=0\)
\(\Leftrightarrow-125x+423=0\)
\(\Leftrightarrow-125x=-423\)
hay \(x=\frac{423}{125}\)
Vậy: \(x=\frac{423}{125}\)
d) Ta có: \(14\frac{1}{2}-\frac{2\left(x+3\right)}{5}=\frac{3x}{2}-\frac{2\left(x-7\right)}{3}\)
\(\Leftrightarrow\frac{435}{30}-\frac{12\left(x+3\right)}{30}-\frac{45x}{30}+\frac{20\left(x-7\right)}{30}=0\)
\(\Leftrightarrow435-12x-36-45x+20x-140=0\)
\(\Leftrightarrow-37x+259=0\)
\(\Leftrightarrow-37x=-259\)
hay \(x=7\)
Vậy: x=7
a) \(3x\left(x-1\right)=x^2-2x+1\)
\(\Leftrightarrow3x\left(x-1\right)=\left(x-1\right)^2\Leftrightarrow\left(x-1\right)\left(x-1-3x\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{2}\end{matrix}\right.\)
b) \(\Leftrightarrow x^3-7x^2+14x-8=0\)
\(\Leftrightarrow x^3-2x^2-5x^2+10x+4x-8=0\)
\(\Leftrightarrow x^2\left(x-2\right)-5x\left(x-2\right)+4\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2-5x+4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=4\end{matrix}\right.\)
c) \(3x^2=4x\Leftrightarrow x\left(3x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{3}\end{matrix}\right.\)
d) \(\Leftrightarrow x^2-6x-7=0\)
\(\Leftrightarrow x^2-6x+9-16=0\)
\(\Leftrightarrow\left(x-3\right)^2-16=0\Leftrightarrow\left(x-7\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-1\end{matrix}\right.\)
Giá trị nguyên x<0 để gt của đa thức :
\(A=12x^3-7x^2-14x+14\) chia hết cho gt của đa thức \(B=4x-5\)
\(A=3x^2\left(4x-5\right)+2x\left(4x-5\right)-\left(4x-5\right)+9\)
\(A=\left(4x-5\right)\left(3x^2+2x-1\right)+9\)
vậy 4x-5 là ước của 9.
x<0=> 4x-5=-9=> x=-1
p/s: Cách tốt nhất để nó không xuất hiện khi nhất chưa ai giải
a) \(\frac{x+\frac{2\left(3-x\right)}{5}}{14}-\frac{5x-4\left(x-1\right)}{24}=\frac{7x+2+\frac{9-3x}{5}}{12}+\frac{2}{3}\)
<=> \(\frac{5x+2\left(3-x\right)}{70}-\frac{5x-4\left(x-1\right)}{24}=\frac{35x+10+9-3x}{60}+\frac{2}{3}\)
<=> \(12\left(5x+6-2x\right)-35\left(5x-4x+4\right)\)
<=> \(14\left(35x+10+9-3x\right)+280.2\) <=> \(12\left(3x+6\right)-35\left(x+4\right)\)
<=> \(14\left(32x+19\right)+560\)
<=> \(36x+72-35x-140=448x+226+560\)
<=> \(-447x=894\)
<=> x = -2
\(\Leftrightarrow7x^2\left(x+2\right)-7\left(x+2\right)\)
\(\Leftrightarrow\left(7x-7\right)\left(x+2\right)\)
suy ra hai trường hợp :
1 : \(7x-7=0\)
\(x=1\)
2: \(x+2=0\)
\(x=-2\)
ĐKXĐ: \(x\ne\pm2\)
\(\frac{x-2}{x+2}\left(\frac{5x+10}{7x-14}+\frac{x-2}{3x-6}\right)=\frac{x-2}{x+2}\left(\frac{5x+10}{7\left(x-2\right)}+\frac{x-2}{3\left(x-2\right)}\right)\)
\(=\frac{x-2}{x+2}\left(\frac{3\left(5x+10\right)}{21\left(x-2\right)}+\frac{7\left(x-2\right)}{21\left(x-2\right)}\right)=\frac{x-2}{x+2}\left(\frac{15x+30}{21\left(x-2\right)}+\frac{7x-14}{21\left(x-2\right)}\right)\)
\(=\frac{x-2}{x+2}\times\frac{22x+16}{21\left(x-2\right)}=\frac{22x+16}{x+2}\)
Ta có phép chia
Chọn đáp án B.