x) \(\left|2x-5\right|=x+1\)

\(\Rightarrow\left[{}\begin{matrix}2x-5=x+1\\2x-5=-x-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x-x=1+5\\2x+x=-1+5\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=6\\3x=4\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=6\\x=\dfrac{4}{3}\end{matrix}\right.\)

y) \(\left|3x-2\right|-1=x\)

\(\Leftrightarrow\left|3x-2\right|=x+1\)

\(\Rightarrow\left[{}\begin{matrix}3x-2=x+1\\3x-2=-x-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}3x-x=1+2\\3x+x=-1+2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=3\\4x=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{1}{4}\end{matrix}\right.\)

z) \(\left|3x-7\right|=2x+1\)

\(\Rightarrow\left[{}\begin{matrix}3x-7=2x+1\\3x-7=-2x-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}3x-2x=1+7\\3x+2x=-1+7\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=8\\5x=6\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=8\\x=\dfrac{6}{5}\end{matrix}\right.\)

v) \(\left|2x-1\right|+1=x\)

\(\Leftrightarrow\left|2x-1\right|=x-1\)

\(\Rightarrow\left[{}\begin{matrix}2x-1=x-1\\2x-1=-x+1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x-x=-1+1\\2x+x=1+1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\3x=2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)

a) \(11+x=-12\)

\(\Leftrightarrow x=-12-11\)

\(\Leftrightarrow x=-23\)

b) \(2-x=17-\left(-5\right)\)

\(\Leftrightarrow2-x=17+5\)

\(\Leftrightarrow2-x=22\)

\(\Leftrightarrow x=2-22\)

\(\Leftrightarrow x=-20\)

c) \(x-12=\left(-9\right)-15\)

\(\Leftrightarrow x-12=-24\)

\(\Leftrightarrow x=-24+12\)

\(\Leftrightarrow x=-12\)

d) \(\left|x\right|-7=9\)

\(\Leftrightarrow\left|x\right|=9+7=16\)

\(\Rightarrow\left[{}\begin{matrix}x=16\\x=-16\end{matrix}\right.\)

e) \(9-25=\left(7-x\right)-\left(25+7\right)\)

\(\Leftrightarrow-16=\left(7-x\right)-32\)

\(\Leftrightarrow-16+32=7-x\)

\(\Leftrightarrow16=7-x\)

\(\Leftrightarrow x=7-16=-9\)

a) - Khi \(x=1\), thay vào ta có:

\(2x^2-8x=2.1^2-8.1=2-8=-6\)

- Khi \(x=\dfrac{1}{2}\) thay vào ta có:

\(2.\left(\dfrac{1}{2}\right)^2-8.\dfrac{1}{2}=2.\dfrac{1}{4}-4=\dfrac{1}{2}-4=\dfrac{-7}{2}\)

b) Thay vào ta có:

\(3.\left(-\dfrac{1}{3}\right)^2+1=3.\dfrac{1}{9}+1=\dfrac{1}{2}+1=\dfrac{3}{2}\)

c) \(\sqrt{x}=\dfrac{1}{2}\Rightarrow x=\dfrac{1}{4}\)

Thay vào ta có:

\(2.\left(\dfrac{1}{4}\right)^2-5.\dfrac{1}{4}+2=2.\dfrac{1}{16}-\dfrac{5}{4}+2=\dfrac{1}{4}-\dfrac{5}{4}+2=-1+2=1\)

a) \(\left|9+x\right|=2x\)

\(\Rightarrow\left[{}\begin{matrix}9+x=2x\\9+x=-2x\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x-x=9\\-2x-x=9\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=9\\-3x=9\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=9\\x=-3\end{matrix}\right.\)

b) \(\left|5x\right|-3x=2\)

\(\Leftrightarrow\left|5x\right|=2+3x\)

\(\Rightarrow\left[{}\begin{matrix}5x=2+3x\\5x=-2-3x\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}5x-3x=2\\5x+3x=-2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=2\\8x=-2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{4}\end{matrix}\right.\)

c) \(\left|x+6\right|-9=2x\)

\(\Leftrightarrow\left|x+6\right|=2x+9\)

\(\Rightarrow\left[{}\begin{matrix}x+6=2x+9\\x+6=-2x-9\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x-2x=9-6\\x+2x=-9-6\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}-3x=3\\3x=-15\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=-5\end{matrix}\right.\)

d) \(\left|2x-3\right|+x=21\)

\(\Leftrightarrow\left|2x-3\right|=21-x\)

\(\Rightarrow\left[{}\begin{matrix}2x-3=21-x\\2x-3=-21+x\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x+x=21+3\\2x-x=-21+3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}3x=24\\x=-18\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-18\end{matrix}\right.\)

Xóa hoặc đánh dấu là spam.

a) \(-3⋮x+1\)

\(\Rightarrow x+1\inƯ\left(-3\right)=\left\{-1;1;-3;3\right\}\)

Ta có bảng sau:

KL: Vậy...

b) \(5⋮x+2\)

\(\Rightarrow x+2\inƯ\left(5\right)=\left\{-1;1;-5;5\right\}\)

Ta có bảng sau:

KL: Vậy...

c) \(x+5⋮x+1\)

\(\Leftrightarrow x+1+4⋮x+1\)

Vì \(x+1⋮x+1\) nên \(4⋮x+1\Rightarrow x+1\inƯ\left(4\right)=\left\{-1;1;-4;4\right\}\)

Ta có bảng sau:

KL: Vậy...

d) \(x-2⋮x+3\)

\(\Leftrightarrow x+3-5⋮x+3\)

Vì \(x+3⋮x+3\) nên \(-5⋮x+3\Rightarrow x+3\inƯ\left(-5\right)=\left\{-1;1;-5;5\right\}\)

Ta có bảng sau:

KL: Vậy...

e) \(2x+3⋮x-1\)

\(\Leftrightarrow2\left(x-1\right)+5⋮x-1\)

Vì \(2\left(x-1\right)⋮x-1\) nên \(5⋮x-1\Rightarrow x-1\inƯ\left(5\right)=\left\{-1;1;-5;5\right\}\)

Ta có bảng sau:

KL: Vậy...

Mấy bài này không đáng để đăng lên đâu.

Đọc kĩ đề bài trước khi đăng. Đây phải là dạng cơ bản nhất của tìm x.

a) \(15.3-x=2\left(x-4\right)\)

\(\Leftrightarrow15-x=2x-8\)

\(\Leftrightarrow15+8=2x+x\)

\(\Leftrightarrow23=3x\)

\(\Leftrightarrow x=\dfrac{23}{3}\)

b) \(10-\left|3-x\right|=6.\left(-7\right)\)

\(\Leftrightarrow10-\left|3-x\right|=-42\)

\(\Leftrightarrow\left|3-x\right|=10+42\)

\(\Leftrightarrow\left|3-x\right|=52\)

\(\Leftrightarrow\left[{}\begin{matrix}3-x=52\\3-x=-52\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-49\\x=55\end{matrix}\right.\)

c) \(\left(x-2\right)\left(2y-1\right)=-6\)

\(\Leftrightarrow\left(x-2\right)\left(2y-1\right)=\left(-1\right).6=6.\left(-1\right)=\left(-2\right).3=3.\left(-2\right)=\left(-6\right).1=1.\left(-6\right)=\left(-3\right).2=2.\left(-3\right)\)

Ta có bảng sau:

KL: Vậy...

Anh Ngọc tích đi, giải cả theo cách lập hệ pt đấy

Truong Son is the longest ………………..……….. in my country.