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a)\(x-15\%x=\frac{1}{3}\)
\(x.\left(1-15\%\right)=\frac{1}{3}\)
\(x.\frac{-280}{3}=\frac{1}{3}\)
\(x=\frac{1}{3}:\frac{-280}{3}\)
\(x=\frac{-1}{280}\)
Vậy \(x=\frac{-1}{280}\)
b)\(\frac{4}{5}x-x-\frac{3}{2}x+\frac{6}{5}=\frac{1}{2}-\frac{4}{3}\)
\(-\frac{17}{10}x+\frac{6}{5}=\frac{-5}{6}\)
\(-\frac{17}{10}x=-\frac{5}{6}-\frac{6}{5}\)
\(-\frac{17}{10}x=\frac{-61}{30}\)
\(x=\frac{-61}{30}:\frac{-17}{10}\)
\(x=\frac{61}{51}\)
Vậy \(x=\frac{61}{51}\)
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\(\Rightarrow4x+8-3+6x=7-2x-2\)
\(\Rightarrow4x+6x+2x=-8+3-2\)
\(\Rightarrow12x=-7\)
\(\Rightarrow x=\frac{-7}{12}\)
Vậy \(x=\frac{-7}{12}.\)
bạn hoàng thị ngọc ánh làm sai rồi
\(4.\left(x+2\right)-3\left(1-2x\right)=7-2\left(x+1\right)\)
\(\Rightarrow4x+8-3-6x=7-2x-2\)
\(\Rightarrow4x+2x-6x+8-3=5\)
\(\Rightarrow6x-6x+8=5+3\)
\(\Rightarrow6x-6x=8-8\)
\(\Rightarrow6x-6x=0\)
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\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}\)
\(=\frac{1}{5}\)
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right)\)
\(\Leftrightarrow=\frac{1}{2}x\frac{2}{3}x\frac{3}{4}x\frac{4}{5}\)
\(\Leftrightarrow=\frac{1x2x3x4}{2x3x4x5}\)
\(\Leftrightarrow=\frac{1}{5}\)
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Ta có: A=(1-1/2)...........................
Mà các tử có hiệu bằng 0
suy ra: Phân số có tử bằng 0
suy ra: A=0
Vậy A=0
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x2 - 3x - 2x2 + 2x = -x2 + 5x
x2 - 3x - 2x2 + 2x + x2 - 5x = 0
(x2 - 2x2 + x2) + (-3x + 2x - 5x) = 0
-6x = 0
x = 0
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\(3\left(x-1\right)-2\left(x+2\right)=3\left(x+2\right)-2x\left(2+3x\right)\)
\(\Rightarrow3\left(x-1\right)-3\left(x+2\right)=2\left(x+2\right)-2x\left(2+3x\right)\)
\(\Rightarrow3\left(x-1-x-2\right)=2\left(x+2\right)-2\left(2x+3x^2\right)\)
\(\Rightarrow3\left(-3\right)=2\left(x+2-2x-3x^2\right)\)
\(\Rightarrow-9=2\left(2-x-3x^2\right)\)
\(\Rightarrow2-x-3x^2=-4,5\)
\(\Rightarrow x-3x^2=6,5\)(hình như sai đề)
\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x.\left(x+1\right)}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\)
\(=\frac{1}{2}-\frac{1}{x+1}\)
\(=\frac{x+1}{2.\left(x+1\right)}-\frac{2}{2.\left(x+1\right)}=\frac{\left(x+1\right)-2}{2.\left(x+1\right)}\)