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Nguyễn Trà My
Phần a)
\(3\times\left(\frac{1}{2}-x\right)+\frac{1}{3}=\frac{7}{6}-x\)
\(32-3x+13=76-x\)
\(116-3x=76-x\)
\(116-76=3x-x\)
\(46=2x\)
\(x=46\div2\)
\(x=13\)
Bài 1:
- \(\dfrac{11}{2}x\) + 1 = \(\dfrac{1}{3}x-\dfrac{1}{4}\)
- \(\dfrac{11}{2}\)\(x\) - \(\dfrac{1}{3}\)\(x\) = - \(\dfrac{1}{4}\) - 1
-(\(\dfrac{33}{6}\) + \(\dfrac{2}{6}\))\(x\) = - \(\dfrac{5}{4}\)
- \(\dfrac{35}{6}\)\(x\) = - \(\dfrac{5}{4}\)
\(x=-\dfrac{5}{4}\) : (- \(\dfrac{35}{6}\))
\(x\) = \(\dfrac{3}{14}\)
Vậy \(x=\dfrac{3}{14}\)
Bài 2: 2\(x\) - \(\dfrac{2}{3}\) - 7\(x\) = \(\dfrac{3}{2}\) - 1
2\(x\) - 7\(x\) = \(\dfrac{3}{2}\) - 1 + \(\dfrac{2}{3}\)
- 5\(x\) = \(\dfrac{9}{6}\) - \(\dfrac{6}{6}\) + \(\dfrac{4}{6}\)
- 5\(x\) = \(\dfrac{7}{6}\)
\(x\) = \(\dfrac{7}{6}\) : (- 5)
\(x\) = - \(\dfrac{7}{30}\)
Vậy \(x=-\dfrac{7}{30}\)
1) \(\left|x+\frac{4}{5}\right|+\frac{7}{5}=\frac{3}{5}\)
\(\Rightarrow\left|x+\frac{4}{5}\right|=\frac{3}{5}-\frac{7}{5}\)
\(\Rightarrow\left|x+\frac{4}{5}\right|=\frac{-4}{5}\)
\(x+\frac{4}{5}=\pm\frac{4}{5}\)
\(TH1:x+\frac{4}{5}=\frac{4}{5}\)
\(\Rightarrow x=\frac{4}{5}-\frac{4}{5}=0\)
\(TH2:x+\frac{4}{5}=\frac{-4}{5}\)
\(\Rightarrow x=\frac{-4}{5}-\frac{4}{5}=\frac{-8}{5}\)
Vậy x ∈ {0; \(\frac{-8}{5}\)}
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a) 3/2.|x - 5/3| - 4/5 = 4/3.|x - 5/3| + 1
<=> 3/2.|x - 5/3| = 4/3.|x - 5/3| + 1 + 4/5
<=> 3/2.|x - 5/3| = 9/5 + 4|x - 5/3|/3
<=> 3/2.|x - 5/3| - 4.|x - 5/3|/3 = 9/5
<=> |x - 5/3|/6 = 9/5
<=> |x - 5/3| = 9/5.6
<=> |x - 5/3| = 54/5
<=> x - 5/3 = 54/5 hoặc x - 5/3 = -54/5
x = 54/5 + 5/3 x = -54/5 - 5/3
x = 187/15 x = -137/15
b) 2.|3x + 1| = 1/3.|3x + 1| + 5
<=> 2.|3x + 1| - 1/3.|3x + 1| = 5
<=> 5/3.|3x + 1| = 5
<=> 5.|3x + 1| = 5.3
<=> 5.|3x + 1| = 15
<=> |3x + 1| = 15 : 5
<=> |3x + 1| = 3
3x + 1 = 3 hoặc 3x + 1 = -3
3x = 3 - 1 3x = -3 - 1
3x = 2 3x = -4
x = 2/3 x = -4/3
=> x = 2/3 hoặc x = -4/3
c) làm tương tự câu a) mình hơi lời
Làm câu c) cho
\(\frac{1}{4}-\frac{5}{2}\left|3x-\frac{1}{5}\right|=\frac{2}{3}\left|3x-\frac{1}{5}\right|-\frac{2}{3}\)
\(\Leftrightarrow\frac{1}{4}+\frac{2}{3}=\frac{2}{3}\left|3x-\frac{1}{5}\right|+\frac{5}{2}\left|3x-\frac{1}{5}\right|\)
\(\Leftrightarrow\frac{3}{12}+\frac{8}{12}=\left|3x-\frac{1}{5}\right|\left(\frac{2}{3}+\frac{5}{2}\right)\)
\(\Leftrightarrow\left|3x-\frac{1}{5}\right|\left(\frac{4}{6}+\frac{15}{6}\right)=\frac{11}{12}\)
\(\Leftrightarrow\frac{19}{6}\left|3x-\frac{1}{5}\right|=\frac{11}{12}\)
\(\Leftrightarrow\left|3x-\frac{1}{5}\right|=\frac{11}{12}.\frac{6}{19}\)
\(\Leftrightarrow\left|3x-\frac{1}{5}\right|=\frac{11}{38}\)
\(\Leftrightarrow\orbr{\begin{cases}3x-\frac{1}{5}=\frac{11}{38}\\3x-\frac{1}{5}=-\frac{11}{38}\end{cases}}\)
Giải tiếp nha
HOC24 có câu rất hay :Người hay giúp bạn khác trả lời bài tập sẽ trở thành học sinh giỏi. Người hay hỏi bài thì không. Còn bạn thì sao? đúng tính bà đó . Lên lớp đừng đập nha :)
a) 3 . ( 1/2 - x ) + 1/3 = 7/6 - x
=> 3/2 - 3x + 1/3 = 7/6-x
=> -3x +x=7/6 - 3/2 - 1/3
=> -2x = -2/3
=> x=-2/3 : (-2) = 1/3
hết :)
a, (2/5-x) : 4/3 + 1/2 = -4
(2/5-x) : 4/3 = -4 -1/2
(2/5-x) : 4/3 = -8/2 - 1/2
(2/5-x) : 4/3 = -9/2
2/5-x = -9/2 . 4/3
2/5-x = -6
x = 2/5 -(-6)
x = 2/5 + 30/5
x = 32/5
Vậy x = 32/5
b, (-3 + 3/x - 1/3) : (1 + 2/5 + 2/3) = -5/4
(-9/3 - 1/3 + 3/x) : (15/15 + 6/15 + 10/15) = -5/4
(-10/3 + 3/x) : 31/15 = -5/4
-10/3 + 3/x = -5/4 . 31/15
-10/3 + 3/x = -31/12
3/x = -31/12 - (-10/3)
3/x = -31/12 + 40/12
3/x = 9/12
3/x = 3/4
Suy ra x= 4
Vậy x = 4
Chúc bạn học tốt
1: Ta có: \(2x+x\left(x-5\right)=3x^2-x\)
\(\Leftrightarrow2x+x^2-5x-3x^2+x=0\)
\(\Leftrightarrow-2x^2-2x=0\)
\(\Leftrightarrow-2x\left(x+1\right)=0\)
Vì -2≠0
nên \(\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
Vậy: x∈{0;-1}
2) Ta có: \(15-5\left(1-2x\right)=12-x\)
\(\Leftrightarrow15-5+10x-12+x=0\)
\(\Leftrightarrow11x-2=0\)
\(\Leftrightarrow11x=2\)
hay \(x=\frac{2}{11}\)
Vậy: \(x=\frac{2}{11}\)
3) Ta có: \(\frac{2}{3}-\frac{1}{3}\left(x-\frac{3}{2}\right)-\frac{1}{2}\left(2x+1\right)=5\)
\(\Leftrightarrow\frac{2}{3}-\frac{1}{3}x+\frac{1}{2}-x-\frac{1}{2}-5=0\)
\(\Leftrightarrow\frac{-13}{3}-\frac{4}{3}x=0\)
\(\Leftrightarrow\frac{4}{3}x=\frac{-13}{3}\)
hay \(x=\frac{-13}{3}:\frac{4}{3}=\frac{-13}{4}\)
Vậy: \(x=\frac{-13}{4}\)
4) Ta có: \(\left|x-\frac{4}{5}\right|=\frac{3}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{4}{5}=\frac{3}{5}\\x-\frac{4}{5}=\frac{-3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{5}\\x=\frac{1}{5}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{1}{5};\frac{7}{5}\right\}\)
1. \(2x+x\left(x-5\right)=3x^2-x\)
\(\Leftrightarrow2x+x^2-5x=3x^2-x\)
\(\Leftrightarrow\left(2x-5x+x\right)+\left(x^2-3x^2\right)=0\)
\(\Leftrightarrow-2x-2x^2=0\)
\(\Leftrightarrow-2x\left(1+x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x=0\\1+x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
2. \(15-5\left(1-2x\right)=12-x\)
\(\Leftrightarrow15-5+10x=12-x\)
\(\Leftrightarrow\left(15-5-12\right)+\left(10x+x\right)=0\)
\(\Leftrightarrow-2+11x=0\)
\(\Leftrightarrow11x=2\Leftrightarrow x=\frac{2}{11}\)
3. \(\frac{2}{3}-\frac{1}{3}\left(x-\frac{3}{2}\right)-\frac{1}{2}\left(2x+1\right)=5\)
\(\Leftrightarrow\frac{2}{3}-\frac{1}{3}x+\frac{1}{2}-x-\frac{1}{2}=5\)
\(\Leftrightarrow\left(\frac{2}{3}+\frac{1}{2}-\frac{1}{2}-5\right)-\left(\frac{1}{3}x+x\right)=0\)
\(\Leftrightarrow-\frac{13}{3}-\frac{4}{3}x=0\)
\(\Leftrightarrow-\frac{4}{3}x=\frac{13}{3}\Leftrightarrow x=-\frac{13}{4}\)
4. \(\left|x-\frac{4}{5}\right|=\frac{3}{5}\)
\(\Rightarrow x-\frac{4}{5}=-\frac{3}{5}\) hoặc \(x-\frac{4}{5}=\frac{3}{5}\)
\(TH1:x-\frac{4}{5}=-\frac{3}{5}\Rightarrow x=\frac{1}{5}\)
\(TH2:x-\frac{4}{5}=\frac{3}{5}\Rightarrow x=\frac{7}{5}\)
\(\left(\dfrac{-4}{5}+\dfrac{-3}{2}:x\right)\cdot\dfrac{1}{3}-\dfrac{1}{2}=\dfrac{1}{2}\\ \left(\dfrac{-4}{5}+\dfrac{-3}{2}:x\right)\cdot\dfrac{1}{3}=\dfrac{1}{2}+\dfrac{1}{2}\\ \left(\dfrac{-4}{5}+\dfrac{-3}{2}\right):x\cdot\dfrac{1}{3}=1\\ \dfrac{-4}{5}+\dfrac{-3}{2}:x=1:\dfrac{1}{3}\\ -\dfrac{4}{5}+\dfrac{-3}{2}x=3\\ \dfrac{3}{2}x=\dfrac{-4}{5}-3\\ \dfrac{3}{2}x=-\dfrac{19}{5}\\ x=\dfrac{-19}{5}:\dfrac{3}{2}\\x =-\dfrac{38}{15}\)