20. Tìm x biết : a)| 5/3 x|=|- 1/6 |; b)| 3/4 x- 3/4 |- 3/4 =|- 3/4 |; c)|x+ 3/5 |-|x- 7/3 |=0
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Câu 6: Khôg có cau nào đúng
Câu 7: C
Câu 8: B
Câu 9: B
Câu 10: D
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a: Ta có: \(\dfrac{1}{4}:x=3\dfrac{4}{5}:40\dfrac{8}{15}\)
\(\Leftrightarrow x=\dfrac{1}{4}\cdot\dfrac{\dfrac{608}{15}}{3+\dfrac{4}{5}}\)
\(\Leftrightarrow x=\dfrac{152}{15}:\dfrac{19}{5}=\dfrac{8}{3}\)
b: Ta có: \(\left(x+1\right):\dfrac{5}{6}=\dfrac{20}{3}\)
\(\Leftrightarrow x+1=\dfrac{50}{9}\)
hay \(x=\dfrac{41}{9}\)
c: Ta có: \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\)
\(\Leftrightarrow x^2-1=63\)
\(\Leftrightarrow x^2=64\)
hay \(x\in\left\{8;-8\right\}\)
c. \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\)
\(7.9=\left(x-1\right).\left(x+1\right)\)
\(63=x^2-1\)
\(x^2=63+1\)
\(x^2=64\)
\(x^2=8^2\)
\(x=8\)
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a. |x-1/5|=|-5/2|-|1/4-2/3|
=> |x-1/5|=5/2-5/12
=> |x-1/5|=25/12
+) x-1/5=25/12
=> x=137/60
+) x-1/5=-25/12
=> x=-113/60
Vậy x \(\in\){-113/60 ; 137/60}
b. (x-0,5).(x2-1,96)=0
+) x-0,5=0
=> x=0,5
+) x2-1,96=0
=> x2=1,96
=> x2=(7/5)2=(-7/5)2
=> x \(\in\left\{-\frac{7}{5};\frac{7}{5}\right\}\)
Vậy \(x\in\left\{-\frac{7}{5};0,5;\frac{7}{5}\right\}\).
c. 5/6 : x = 20 : 3
=> 5/6 : x = 20/3
=> x = 5/6 : 20/3
=> x=1/8
Vậy x=1/8.
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-( -x + 13 - 142 ) + 18 = 55
-( -x - 129) + 18 = 55
x + 129 + 18 = 55
x + 147 = 55
x = 55 - 147
x = -92
b, 25 - 3.(6 -x ) = 22
3(6-x) = 25 - 22
3(6-x) = 3
6 - x = 1
x =5
c, [ ( 2x - 11 ) :3 +1] .5 = 20
( 2x - 11) : 3 + 1 = 20 : 5
(2x - 11) : 3 + 1 = 4
( 2x - 11) : 3 = 4 - 1
(2x - 11 ) : 3 = 3
2x - 11 = 3 x 3
2x - 11 = 9
2x = 9 + 11
2x = 20
x = 10
d, 3(x+5) - x - 11 = 24
3(x+5) - x = 24 +11
3x + 15 - x = 35
2x = 35 - 15
2x = 20
x = 10
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\(a,\Leftrightarrow x-1=4\Leftrightarrow x=5\\ b,\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\3x+1=4x-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\x=4\left(tm\right)\end{matrix}\right.\Leftrightarrow x=4\\ c,ĐK:x\ge-5\\ PT\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\\ \Leftrightarrow3\sqrt{x+5}=6\\ \Leftrightarrow\sqrt{x+5}=3\\ \Leftrightarrow x+5=9\\ \Leftrightarrow x=4\left(tm\right)\)
\(d,\Leftrightarrow\sqrt{\left(x-2\right)^2}=\sqrt{\left(\sqrt{5}+1\right)^2}\\ \Leftrightarrow\left|x-2\right|=\sqrt{5}+1\\ \Leftrightarrow\left[{}\begin{matrix}x-2=\sqrt{5}+1\\2-x=\sqrt{5}+1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{5}+3\\x=1-\sqrt{5}\end{matrix}\right.\)
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1
\(\left(x-2\right):2.3=6\)
\(\Leftrightarrow\left(x-2\right):2=2\)
\(\Leftrightarrow\left(x-2\right)=4\)
\(\Leftrightarrow x=4+2=6\)
c) ta có
\(\left[\left(2x+1\right)+1\right]m:2=625\)
\(\Leftrightarrow\left[\left(2x+1\right)+1\right]\left\{\left[\left(2x+1\right)-1\right]:2+1\right\}=1250\)
\(\Leftrightarrow\left(2x+1\right)^2+1-1:2+1=1250\)
\(\Leftrightarrow\left(2x+1\right)^2+1-2+1=1250\)
\(\Leftrightarrow\left(2x+1\right)^2+1-2=1249\)
\(\Leftrightarrow\left(2x+1\right)^2+1=1251\)
\(\Leftrightarrow\left(2x+1\right)^2=1250\)
...
2
\(\left(x-\frac{1}{2}\right).\frac{5}{3}=\frac{7}{4}-\frac{1}{2}\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right).\frac{5}{3}=\frac{5}{4}\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)=\frac{5}{4}:\frac{5}{3}\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)=\frac{5}{4}.\frac{3}{5}\)
\(\Leftrightarrow x-\frac{1}{2}=\frac{3}{4}\)
\(\Leftrightarrow x=\frac{3}{4}+\frac{1}{2}=\frac{5}{4}\)
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a)\(\frac{X}{5}=\frac{5}{6}+\frac{-19}{30}\)
\(\frac{X}{5}=\frac{1}{5}\)
Vậy \(X=1\)
b)\(X-\frac{41}{5}=\frac{-2}{3}\)
\(X=\frac{-2}{3}+\frac{41}{5}\)
\(X=\frac{113}{15}\)
Vậy \(X=\frac{113}{15}\)
c)\(\frac{31}{5}-X=\frac{11}{3}+\frac{7}{10}\)
\(\frac{31}{5}-X=\frac{131}{30}\)
\(X=\frac{31}{5}-\frac{131}{30}\)
\(X=\frac{11}{6}\)
Vậy \(X=\frac{11}{6}\)
d)\(\frac{9}{X}=\frac{2}{5}+\frac{-7}{20}\)
\(\frac{9}{X}=\frac{1}{20}\)
\(X=9:\frac{1}{20}\)
\(X=180\)
Vậy \(X=180\)
Hc tốt
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Làm
a) x/20 - 1/5 = 1/4
x/20 = 1/4 + 1/5
x /20 = 5/20 + 4/20
x/20 = 9/20
=> x = 9
b) Phần b thiếu đề bài ạ nên mk k giải đc phần b
c) | x - 1/2 | = 5/6
x - 1/2 = 5/6 x = 5/6 + 1/2
=> { => {
x - 1/2 = -5/6 x = -5/6 + 1/2
x = 4/3
=>
x = -1/3
Vậy x = 4/3 hoặc x = -1/3
HỌC TỐT
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a, \(\frac{x}{5}=\frac{-12}{20}\)
\(x=\frac{5.\left(-20\right)}{20}\)
\(x=-5\)
b, 3x - 35 = -45 - 2x
3x + 2x = -45 + 35
5x = -10
x = (-10) : 5
x = -2
c, \(\frac{-1}{6}-x=\frac{5}{12}+\frac{-2}{3}\)
\(\frac{-1}{6}-x=\frac{-1}{4}\)
\(-x=\frac{-1}{6}-\frac{-1}{4}\)
\(-x=\frac{1}{12}\)
\(x=\frac{-1}{12}\)
a: \(\Leftrightarrow\left|\dfrac{5}{3}x\right|=\dfrac{1}{6}\)
\(\Leftrightarrow\left[{}\begin{matrix}x\cdot\dfrac{5}{3}=\dfrac{1}{6}\\x\cdot\dfrac{5}{3}=-\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}:\dfrac{5}{3}=\dfrac{3}{30}=\dfrac{1}{10}\\x=-\dfrac{1}{10}\end{matrix}\right.\)
b: \(\Leftrightarrow\left|\dfrac{3}{4}x-\dfrac{3}{4}\right|=\dfrac{3}{4}+\dfrac{3}{4}=\dfrac{3}{2}\)
\(\Leftrightarrow\left|x-1\right|=\dfrac{3}{2}:\dfrac{3}{4}=2\)
=>x-1=2 hoặc x-1=-2
=>x=3 hoặc x=-1
c: \(\Leftrightarrow\left|x+\dfrac{3}{5}\right|=\left|x-\dfrac{7}{3}\right|\)
\(\Leftrightarrow x+\dfrac{3}{5}=\dfrac{7}{3}-x\)
=>2x=44/15
hay x=22/15