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Bài 1:
a,\(0,75+\frac{9}{17}-1\frac{4}{5}-\frac{26}{17}-2\frac{4}{5}\)
\(=\frac{3}{4}+\left(\frac{9}{17}-\frac{26}{17}\right)-\left(1\frac{4}{5}+2\frac{4}{5}\right)\)
\(=\frac{3}{4}-1-\frac{23}{5}\)
\(=\frac{15}{20}-\frac{20}{20}-\frac{92}{20}=\frac{-97}{20}\)
Bài 2:
a, \(\left(2x+\frac{3}{4}\right)-\frac{10}{3}=\frac{-13}{3}\)
\(2x+\frac{3}{4}=\frac{-13}{3}+\frac{10}{3}\)
\(2x+\frac{3}{4}=-1\)
\(2x=-1-\frac{3}{4}\)
\(2x=\frac{-7}{4}\)
x = -7/8
b, 3,2x - 2,7x + 8,5 = 6
x(3,2 - 2,7) = -2,5
0,5x = -2,5
x = -5
a, 1,5 +|2x - 2/3| = 3/2
|2x - 2/3| = 3/2 - 1,5
|2x - 2/3| = 0
<=> 2x - 2/3 = 0
<=> 2x = 0 + 2/3
<=> 2x = 2/3
<=> x = 2/3 : 2
<=> x = 1/3
Vậy x = 1/3
b, 3/4 - |1/4 - x| = 5/8
|1/4 - x| = 3/4 - 5/8
|1/4 - x| = 1/8
<=> 1/4 - x = 1/8
1/4 - x = /1/8
<=> x = 1/4 - 1/8
x = 1/4 - ( -1/8)
<=> x = 1/8
x = 3/8
Vậy x thuộc { 1/8 ; 3/8 }
1) 1/3x-1/2 = -1/2x+1/8-2/3
1/3x+1/2x = 1/2+1/8-2/3
5/6x = -1/24
vậy x =-1/20
c: Ta có: \(\dfrac{2}{5}\cdot\left[\left(\dfrac{3}{5}\right)^2:\left(-\dfrac{1}{5}\right)^2-7\right]\cdot\left(1000\right)^0\cdot\left|-\dfrac{11}{15}\right|\)
\(=\dfrac{2}{5}\cdot\left(\dfrac{9}{25}:\dfrac{1}{25}-7\right)\cdot1\cdot\dfrac{11}{15}\)
\(=\dfrac{2}{5}\cdot\dfrac{11}{15}\cdot2\)
\(=\dfrac{44}{75}\)
A. 2.\(|3x+1|\)=\(\frac{3}{4}\)-\(\frac{5}{8}\)
2.\(|3x+1|\)=1/8
\(|3x+1|\)=1/8:2
\(|3x+1|\)=1/16
TH1 : 3x+1=1/16
3x=1/16-1
3x=-15/16
x=-15/16:3
x=-5/16
a,\(\frac{3}{4}-2.\left|3x+1\right|=\frac{5}{8}\)
\(\Rightarrow2.\left|3x+1\right|=\frac{3}{4}-\frac{5}{8}=\frac{6}{8}-\frac{5}{8}=\frac{1}{8}\)
\(\Rightarrow\left|3x+1\right|=\frac{1}{8}.\frac{1}{2}=\frac{1}{16}\)
\(\Rightarrow\orbr{\begin{cases}3x+1=\frac{1}{16}\\3x+1=\frac{-1}{16}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}3x=\frac{1}{16}-1=\frac{-15}{16}\\3x=\frac{-1}{16}-1=\frac{-17}{16}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{-15}{16}.\frac{1}{3}=\frac{-5}{16}\\x=\frac{-17}{16}.\frac{1}{3}=\frac{-17}{48}\end{cases}}\)
Vậy....
b,\(\left|3x+2\right|-\left|x-3\right|=\frac{7}{2}\left(1\right)\)
Ta có bảng xét dấu
x | \(\frac{-2}{3}\) 3 |
3x+2 | - 0 + | + |
x-3 | - | - 0 + |
Nếu x<\(\frac{-2}{3}\) thì \(\left|3x+2\right|-\left|x-3\right|\) \(=-3x-2-3+x\)
\(=-2x-5\)
Từ (1) \(\Rightarrow-2x-5=\frac{7}{2}\)
\(\Rightarrow-2x=\frac{7}{2}+5=\frac{17}{2}\)
\(\Rightarrow x=\frac{17}{2}\cdot\frac{-1}{2}=\frac{-17}{4}\)(thỏa mãn x<\(\frac{-2}{3}\)
Nếu \(\frac{-2}{3}\le x\le3\)thì \(\left|3x+2\right|-\left|x-3\right|=3x+2-\left(3-x\right)\)
\(=3x+2-3+x\)
\(=2x-1\)
Từ (1)\(\Rightarrow\)\(2x-1=\frac{7}{2}\)
\(\Rightarrow2x=\frac{9}{2}\)
\(\Rightarrow x=\frac{9}{4}\)(thỏa mãn......
Còn trưonwfg hợp cuối bạn tự làm nốt nhé
a)
\(\left|x+1\right|\ge0\forall x\Rightarrow2x\ge0\forall x\Rightarrow x\ge0\forall x\)
=> x + 1 = 2x
=> 2x - x = 1
=> x = 1
P.s : đợi chút mấy câu kia
b)
Nếu \(x\ge0\)thì :
x - 3 = x - 4
x - x = -4 + 3
0.x = -1 ( loại )
Nếu \(x\le0\)thì :
x - 3 = -x + 4
x + x = 4 + 3
2x = 7
x = 7/2 ( tm )
Vậy x = 7/2
Bài 4:
b: Ta có: \(2x\left(x-\dfrac{1}{4}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{4}\end{matrix}\right.\)
a, (3 + x)2 = 4
(3 + x)2 = 22 = (-2)2
TH1: 3 + x = 2
x = 2 - 3
x = -1
TH2: 3 + x = -2
x = -2 - 3
x = -2 + (-3)
x = -5
Vậy x ϵ {-1; -5}
b, (2x + 1) 3 = -8
(2x + 1)3 = (-2)3
2x + 1 = -2
2x = -2 - 1
2x = -2 + (-1)
2x = -3
x = \(\dfrac{-3}{2}\)
có ai ko