Tìm x,y:
I\(\frac{1}{2}-\frac{1}{3}+x\)I=-IyI-\(\frac{1}{4}\)
Ix-yI+Iy+\(\frac{5}{17}\)I=0
Ix-\(\frac{3}{5}\)I<\(\frac{1}{3}\)
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a)\(\left|x+\frac{1}{5}\right|-4=-2\)
\(\Rightarrow\left|x+\frac{1}{5}\right|=2\)
\(\Rightarrow x+\frac{1}{5}=2\) hoặc \(-2\)
Xét \(x+\frac{1}{5}=2\Leftrightarrow x=\frac{9}{5}\)
Xét \(x+\frac{1}{5}=-2\Leftrightarrow x=-\frac{11}{5}\)
e)
=> (x-2) . (x+7) = ( x-1 ) . ( x +4)
=> x2 +7x - 2x -14 = x2 - x + 4x - 4
x2 + 5x - 14 = x2 + 3x - 4
=> 5x - 14 = 3x - 4
=> 5x - 3x = 14-4
=> 2x = 10 => x = 10 : 2 => x = 5
c)
=>( x-1) . 7 = ( x + 5 ) . 6
=> 7x - 7 = 6x + 30
=> 7x - 6x= 30 + 7
=> x = 37
a,x=\(\frac{5}{2}\)
b,x=\(\frac{13}{176}\)
c,x=37
d, x=\(\frac{12}{5}\)
e, x=5
1) \(\left(x-2\right)\left(\frac{x+1}{3}-x+1\right)=0\)
\(\Leftrightarrow\frac{x\left(x+1\right)}{3}-x^2+x-\frac{2\left(x+1\right)}{3}+2x-2=0\)
\(\Leftrightarrow\frac{x\left(x+1\right)}{3}-x^2+3x-\frac{2\left(x+1\right)}{3}-2=0\)
\(\Leftrightarrow x\left(x+1\right)-3x^2+9x-2\left(x+1\right)-6=0\)
\(\Leftrightarrow x^2+x-3x^2+9x-2x-2-6=0\)
\(\Leftrightarrow-2x^2+8x-8=0\)
\(\Leftrightarrow-2\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow-2.\left(x^2-2.x.2+2^2\right)=0\)
\(\Leftrightarrow-2\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(x-2\right)^2=0\)
\(\Leftrightarrow x-2=0\)
\(\Leftrightarrow x=2\)
Vậy nghiệm của phương trình là: {2}
2) \(\left(3x+4x\right)\left(\frac{x}{2}-x-\frac{3x}{5}+1\right)=0\)
\(\Leftrightarrow7x\left(\frac{x}{2}-x-\frac{3x}{5}+1\right)=0\)
\(\Leftrightarrow7x\left(-\frac{11x}{10}+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}7x=0\\-\frac{11x}{10}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{11}{10}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{10}{11}\end{cases}}\)
Vậy: nghiệm của phương trình là: \(\left\{0;\frac{10}{11}\right\}\)
3) \(\left|x-1\right|=x^2-x\)
\(\Leftrightarrow x-1=x^2-x\)
\(\Leftrightarrow1=x^2-x-x\)
\(\Leftrightarrow1=x^2\)
\(\Leftrightarrow x^2=1\)
\(\Rightarrow x=\pm1\)
Vậy nghiệm phương trình là: {1; -1}
4) \(\left|x^2-3x+1\right|=2x-3\)
\(\Leftrightarrow\orbr{\begin{cases}x^2-3x+1=2x-3\\x^2-3x+1=-\left(2x-3\right)\end{cases}}\)
Xét trường hợp này rồi làm tiếp, dễ rồi :))
1)\(\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{4}=\frac{x+1}{5}+\frac{x+1}{6}\)
\(\Leftrightarrow\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{4}-\frac{x+1}{5}-\frac{x+1}{6}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\right)=0\)
\(\Leftrightarrow\left(x+1\right)=0\).Do \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\ne0\)
\(\Leftrightarrow x=-1\)
\(1a,\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)
\(\Leftrightarrow\frac{3\left(2x+1\right)^2}{15}-\frac{5\left(x-1\right)^2}{15}=\frac{7x^2-14x-5}{15}\)
\(\Leftrightarrow\frac{12x^2+12x+3}{15}-\frac{5x^2-10x+5}{15}=\frac{7x^2-14x-5}{15}\)
\(\Leftrightarrow12x^2+12x+3-5x^2+10x-5=7x^2-14x-5\)
\(\Leftrightarrow36x=-3\)
\(x=-\frac{1}{12}\)
Vậy ................
\(b,\frac{7x-1}{6}+2x=\frac{16-x}{5}\)
\(\Leftrightarrow\frac{5\left(7x-1\right)}{30}+\frac{30.2x}{30}=\frac{6\left(16-x\right)}{30}\)
\(\Leftrightarrow35x-5+60x=96-6x\)
\(\Leftrightarrow101x=101\)
\(\Leftrightarrow x=1\)
Vậy ....................
\(\frac{3}{4}-\frac{5}{6}\le\frac{x}{12}< 1-\left(\frac{2}{3}-\frac{1}{4}\right)\)
\(\Leftrightarrow-\frac{1}{12}\le\frac{x}{12}< \frac{7}{12}\)
=> x \(\in\) {-1;0;1;2;3;4;5;6}
\(\frac{3}{4}-\frac{5}{6}\le\frac{x}{12}< 1-\left(\frac{2}{3}-\frac{1}{4}\right)\)
\(\Leftrightarrow\)\(\frac{9-10}{12}\le\frac{x}{12}< 1-\left(\frac{8-3}{12}\right)\)
\(\Leftrightarrow\)\(-\frac{1}{12}\le\frac{x}{12}< \frac{7}{12}\)
\(\Leftrightarrow-1\le x< 7\)
Mà x nguyên
=>x={-1;0;1;2;3;4;5;6}
\(\left|x-y\right|+\left|y+\frac{5}{17}\right|=0\)
\(\Leftrightarrow\left|x-y\right|=\left|y+\frac{5}{17}\right|=0\)
\(\Leftrightarrow x=y=-\frac{5}{17}\)