x+3/5=x-4/10
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\(a,\Leftrightarrow\left|x+\dfrac{2}{5}\right|=\dfrac{7}{4}\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{2}{5}=\dfrac{7}{4}\left(x\ge-\dfrac{2}{5}\right)\\x+\dfrac{2}{5}=-\dfrac{7}{4}\left(x< -\dfrac{2}{5}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{27}{20}\left(tm\right)\\x=-\dfrac{43}{20}\left(tm\right)\end{matrix}\right.\)
\(b,\Leftrightarrow\left|x-\dfrac{13}{10}\right|=\dfrac{13}{10}\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{13}{10}=\dfrac{13}{10}\left(x\ge\dfrac{13}{10}\right)\\x-\dfrac{13}{10}=-\dfrac{13}{10}\left(x< \dfrac{13}{10}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{13}{5}\left(tm\right)\\x=0\left(tm\right)\end{matrix}\right.\)
\(c,\Leftrightarrow\left|\dfrac{3}{4}-\dfrac{1}{2}x\right|=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{4}-\dfrac{1}{2}x=\dfrac{1}{2}\left(x\le\dfrac{3}{2}\right)\\\dfrac{1}{2}x-\dfrac{3}{4}=\dfrac{1}{2}\left(x>\dfrac{3}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\x=\dfrac{5}{2}\left(tm\right)\end{matrix}\right.\)
\(d,\Leftrightarrow\left|5-2x\right|=4\Leftrightarrow\left[{}\begin{matrix}5-2x=4\left(x\le\dfrac{5}{2}\right)\\2x-5=4\left(x>\dfrac{5}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\x=\dfrac{9}{2}\left(tm\right)\end{matrix}\right.\)
\(đ,\Leftrightarrow\left\{{}\begin{matrix}x-3,5=0\\x-1,3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3,5\\x=1,3\end{matrix}\right.\left(vô.lí\right)\Leftrightarrow x\in\varnothing\)
\(e,\Leftrightarrow\left\{{}\begin{matrix}x-2021=0\\x-2022=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2021\\x=2022\end{matrix}\right.\left(vô.lí\right)\Leftrightarrow x\in\varnothing\)
\(f,\Leftrightarrow\left|x\right|=\dfrac{1}{3}-x\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}-x\left(x\ge0\right)\\x=x-\dfrac{1}{3}\left(x< 0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\left(tm\right)\\0x=-\dfrac{1}{3}\left(vô.lí\right)\end{matrix}\right.\Leftrightarrow x=\dfrac{1}{6}\)
\(g,\Leftrightarrow\left[{}\begin{matrix}x-2=x\left(x\ge2\right)\\2-x=x\left(x< 2\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}0x=2\left(vô.lí\right)\\x=1\left(tm\right)\end{matrix}\right.\Leftrightarrow x=1\)
Đề trước đó:
(x-7)(x+1)-(x-3)^2=(3x-5)(3x+5)-(3x+1)^2+(x-2)^2-x
<=>x^2+x-7x-7-x^2+6x-9=9x^2-25-9x^2-6x-1+x^2-4x+4-x
<=>x^2-11x-6=0
<=>x^2-2x. 11/2 + 121/4-145/4=0
<=>(x-11/2)^2=145/4
<=>|x-11/2|=căn(145)/2
<=>x=[11+-căn(145)]/2
a: P(x)=0
=>4x-7-x-14=0
=>3x-21=0
=>x=7
b: x^2+x=0
=>x(x+1)=0
=>x=0; x=-1
\(\frac{1}{x}+\frac{2}{x}+\frac{3}{x}+.......+\frac{100}{x}=101\)
\(\Leftrightarrow\frac{1}{x}\left(1+2+3+....+100\right)=101\)
\(\Leftrightarrow\frac{1}{x}.\frac{100.\left(100+1\right)}{2}=101\)
\(\Leftrightarrow\frac{1}{x}.50=1\)
\(\Rightarrow\frac{1}{x}=\frac{1}{50}\)
\(\Rightarrow x=50\)
(-1)+3+(-5)+7+....+x = 600
<=> [(-1) + 3] + [(-5) + 7] .... + [(-x) - 2) + x] = 600
Ta co : 2 + 2 + 2 +.....+ 2 = 600
<=> 1 + 1 + 1 +.....+ 1 = 300
Số dấu ngoặc[ ] la : x−34 +1
=> x−34 +1=300
<=> x−34 =299
<=> x - 3 = 299 . 4 = 1199
Vậy x = 1199
( x + 3 ) / 5 = ( x - 4 ) / 10
=> ( x + 3 ) . 10 = ( x - 4 ) . 5
=> 10x + 30 = 5x - 20
=> 10x + 30 + 20 = 5x
=> 10x + 50 = 5x
=> 5x - 10x = 50
=> -5x = 50
=> x = 50 : -5
=> x = -10
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bạn có thể nói rõ đề bài ko >_< . Trong 2 TH , cái nào đúng ?
Th1 : x + \(\frac{3}{5}\)= x - \(\frac{4}{10}\)
Th2 : \(\frac{x+3}{5}=\frac{x-4}{10}\)