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Tìm x biết
a) (x-1/2)^2=4
b) 10/1/2-(x+1/3)^2=1/1/2
c) (x-1/5)^2+17/25=26/25
d) 1/5/27+(3x-7/9)^3=24/27
![](https://rs.olm.vn/images/avt/0.png?1311)
a) (x - 1/2)2 = 4
<=> (x - 1/2)2 = 22
<=> x - 1/2 = -2; 2
<=> x - 1/2 = 2 hoặc x - 1/2 = -2
x = 2 + 1/2 x = -2 + 1/2
x = 5/2 x = -3/2
=> x = 5/2 hoặc x = -3/2
b) 10/1/2 - (x + 1/3)2 = 1/1/2
<=> -(x + 1/3)2 = 1/1/2 - 10/1/2
<=> -(x + 1/3)2 = 1/2 - 5
<=> -(x + 1/3)2 = -5.2 + 1/2
<=> -(x + 1/3)2 = -9/2
<=> (x + 1/3)2 = 9/2
<=> x + 1/3 = \(\sqrt{\frac{9}{2}}\) hoặc x + 1/3 = \(-\sqrt{\frac{9}{2}}\)
x = \(\frac{3\sqrt{2}}{2}\) - 1/3 x = \(-\frac{3\sqrt{2}}{2}\) -1/3
=> x = \(\frac{3\sqrt{2}}{2}\) - 1/3 hoặc x = \(-\frac{3\sqrt{2}}{2}\) -1/3
c) (x - 1/5)2 + 17/25 = 26/25
<=> (x - 1/5)2 = 26/25 - 17/25
<=> (x - 1/5)2 = (3/5)2
<=> x - 1/5 = -3/5; 3/5
<=> x - 1/5 = 3/5 hoặc x - 1/5 = -3/5
x = 3/5 + 1/5 x = -3/5 + 1/5
x = 4/5 x = -2/5
=> x = 4/5 hoặc x = -2/5
![](https://rs.olm.vn/images/avt/0.png?1311)
`Answer:`
a. \(\frac{17}{2}-\left|2x-\frac{3}{4}\right|=-\frac{7}{4}\)
\(\Leftrightarrow\left|2x-\frac{3}{4}\right|=\frac{17}{2}+\frac{7}{4}\)
\(\Leftrightarrow\left|2x-\frac{3}{4}\right|=\frac{41}{4}\)
\(\Leftrightarrow\orbr{\begin{cases}2x-\frac{3}{4}=\frac{41}{4}\\2x-\frac{3}{4}=-\frac{41}{4}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=\frac{41}{4}+\frac{3}{4}\\2x=-\frac{41}{4}+\frac{3}{4}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=11\\2x=-\frac{19}{2}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=11:2\\x=-\frac{19}{2}:2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{11}{2}\\x=-\frac{19}{4}\end{cases}}\)
b. \(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)
\(\Leftrightarrow\left(x+\frac{1}{5}\right)^2=\frac{26}{25}-\frac{17}{25}\)
\(\Leftrightarrow\left(x+\frac{1}{5}\right)^2=\frac{9}{25}\)
\(\Leftrightarrow\left(x+\frac{1}{5}\right)=\left(\frac{3}{5}\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{5}=\frac{3}{5}\\x+\frac{1}{5}=-\frac{3}{5}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{5}-\frac{1}{5}\\x=-\frac{3}{5}-\frac{1}{5}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{5}\\x=-\frac{4}{5}\end{cases}}\)
c. \(-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\Leftrightarrow-\frac{32}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\Leftrightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{32}{27}-\left(-\frac{24}{27}\right)\)
\(\Leftrightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{8}{27}\)
\(\Leftrightarrow\left(3x-\frac{7}{9}\right)^3=\left(-\frac{2}{3}\right)^3\)
\(\Leftrightarrow3x-\frac{7}{9}=-\frac{2}{3}\)
\(\Leftrightarrow3x=-\frac{2}{3}+\frac{7}{9}\)
\(\Leftrightarrow3x=\frac{1}{9}\)
\(\Leftrightarrow x=\frac{1}{9}:3\)
\(\Leftrightarrow x=\frac{1}{27}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{1}{10}\)
\(\dfrac{2}{3}x=\dfrac{1}{10}+\dfrac{1}{2}=\dfrac{3}{5}\)
\(x=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{9}{10}\)
b) \(\dfrac{39}{7}:x=13\)
\(x=\dfrac{\dfrac{39}{7}}{13}=\dfrac{3}{7}\)
c) \(\left(\dfrac{14}{5}x-50\right):\dfrac{2}{3}=51\)
\(\dfrac{14}{5}x-50=51\cdot\dfrac{2}{3}=34\)
\(\dfrac{14}{5}x=34+50=84\)
\(x=\dfrac{84}{\dfrac{14}{5}}=30\)
d) \(\left(x+\dfrac{1}{2}\right)\left(\dfrac{2}{3}-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\\dfrac{2}{3}-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
e) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\dfrac{5}{12}\)
\(\dfrac{1}{6}x=\dfrac{5}{12}\)
\(x=\dfrac{5}{12}:\dfrac{1}{6}=\dfrac{5}{2}\)
g) \(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\dfrac{11}{5}-\dfrac{3}{7}=-2\)
\(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\cdot\dfrac{11}{5}=-2+\dfrac{3}{7}=-\dfrac{11}{7}\)
\(x\cdot\dfrac{44}{7}+\dfrac{3}{7}=-\dfrac{11}{7}:\dfrac{11}{5}=-\dfrac{5}{7}\)
\(\dfrac{44}{7}x=-\dfrac{5}{7}-\dfrac{3}{7}=-\dfrac{8}{7}\)
\(x=-\dfrac{8}{7}:\dfrac{44}{7}=-\dfrac{2}{11}\)
h) \(\dfrac{13}{4}x+\left(-\dfrac{7}{6}\right)x-\dfrac{5}{3}=\dfrac{5}{12}\)
\(\dfrac{25}{12}x-\dfrac{5}{3}=\dfrac{5}{12}\)
\(\dfrac{25}{12}x=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{25}{12}\)
\(x=1\)
Mỏi tay woa bn làm nốt nha!!
![](https://rs.olm.vn/images/avt/0.png?1311)
d) Ta có: \(32\%-0.25:x=-\dfrac{17}{5}\)
\(\Leftrightarrow0.25:x=\dfrac{8}{25}+\dfrac{17}{5}=\dfrac{93}{25}\)
hay \(x=\dfrac{25}{372}\)
Vậy: \(x=\dfrac{25}{372}\)
e) Ta có: \(\left(x+\dfrac{1}{5}\right)^2+\dfrac{17}{25}=\dfrac{26}{25}\)
\(\Leftrightarrow\left(x+\dfrac{1}{5}\right)^2=\dfrac{9}{25}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{3}{5}\\x+\dfrac{1}{5}=-\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{2}{5};-\dfrac{4}{5}\right\}\)
f) Ta có: \(-\dfrac{32}{27}-\left(3x-\dfrac{7}{9}\right)^3=-\dfrac{24}{27}\)
\(\Leftrightarrow\left(3x-\dfrac{7}{9}\right)^3=\dfrac{-8}{27}\)
\(\Leftrightarrow3x-\dfrac{7}{9}=-\dfrac{2}{3}\)
\(\Leftrightarrow3x=\dfrac{1}{9}\)
hay \(x=\dfrac{1}{27}\)
g) Ta có: \(60\%\cdot x+0.4x+x:3=2\)
\(\Leftrightarrow\dfrac{4}{3}x=2\)
hay \(x=\dfrac{3}{2}\)
Vậy: \(x=\dfrac{3}{2}\)
h) PT \(\Leftrightarrow\left|\dfrac{20}{9}-x\right|=\dfrac{2}{9}\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{20}{9}-x=\dfrac{2}{9}\\x-\dfrac{20}{9}=\dfrac{2}{9}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{22}{9}\end{matrix}\right.\)
Vậy ...
i) PT \(\Leftrightarrow\dfrac{8}{5}+\dfrac{2}{5}x=\dfrac{16}{5}\) \(\Leftrightarrow\dfrac{2}{5}x=\dfrac{8}{5}\) \(\Leftrightarrow x=4\)
Vậy ...
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\\ \left(x+\frac{1}{5}\right)^2=\frac{26}{25}-\frac{17}{25}\\ \left(x+\frac{1}{5}\right)^2=\frac{9}{25}\\ \left|\left(x+\frac{1}{5}\right)\right|=\frac{3}{5}\)
TH1: \(x=\frac{3}{5}-\frac{1}{5}\\ x=\frac{2}{5}\)
TH2: \(\left|\left(x+\frac{1}{5}\right)\right|=-\frac{3}{5}\\ x=-\frac{3}{5}-\frac{1}{5}\\ x=-\frac{4}{5}\)
\(a,\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)
\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\frac{9}{25}\)
\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(\Rightarrow x+\frac{1}{5}=\frac{3}{5}\)
\(\Rightarrow x=\frac{2}{5}\)
\(b,-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\Rightarrow-\frac{32}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{32}{27}+\frac{24}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{8}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=\left(-\frac{2}{3}\right)^3\)
\(\Rightarrow3x-\frac{7}{9}=-\frac{2}{3}\)
\(\Rightarrow3x=-\frac{2}{3}+\frac{7}{9}\)
\(\Rightarrow3x=\frac{1}{9}\)
\(\Rightarrow x=\frac{1}{27}\)
\(c,\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\)
\(\Rightarrow\) \(\left[\begin{array}{nghiempt}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{array}\right.\) \(\Rightarrow\) \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\2x=\frac{2}{3}\end{array}\right.\) \(\Rightarrow\) \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\x=\frac{1}{3}\end{array}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
2,
a) \(315-\left(135-x\right)=215\)
\(\Rightarrow135-x=315-215\)
\(\Rightarrow135-x=100\)
\(\Rightarrow x=135-100\)
\(\Rightarrow x=35\)
b) \(x-320:32=25\cdot16\)
\(\Rightarrow x-10=5^2\cdot4^2\)
\(\Rightarrow x-10=20^2\)
\(\Rightarrow x-10=400\)
\(\Rightarrow x=410\)
c) \(3\cdot x-2018:2=23\)
\(=3\cdot x-1009=23\)
\(\Rightarrow3\cdot x=1032\)
\(\Rightarrow x=1032:3\)
\(\Rightarrow x=344\)
d) \(280-9\cdot x-x=80\)
\(\Rightarrow280-x\cdot\left(9+1\right)=80\)
\(\Rightarrow280-10\cdot x=80\)
\(\Rightarrow10\cdot x=280-80\)
\(\Rightarrow10\cdot x=200\)
\(\Rightarrow x=20\)
e) \(38\cdot x-12\cdot x-x\cdot16=40\)
\(\Rightarrow x\cdot\left(38-12-16\right)=40\)
\(\Rightarrow x\cdot10=40\)
\(\Rightarrow x=40:10\)
\(\Rightarrow x=4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)
\(\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(=>2x+\frac{3}{5}=\frac{3}{5}\)
\(2x=\frac{3}{5}-\frac{3}{5}\)
\(2x=0\)
\(x=0:2\)
\(x=0\)
b) \(\left(3x-1\right).\left(-\frac{1}{2x}+5\right)=0\)
=> \(\left(3x-1\right)=0\)hoặc \(\left(-\frac{1}{2x}+5\right)=0\)hoặc \(\left(3x-1\right)\)và\(\left(-\frac{1}{2x}+5\right)\)cùng bằng 0.
\(\orbr{\begin{cases}3x-1=0\\-\frac{1}{2x}+5=0\end{cases}}=>\orbr{\begin{cases}3x=1\\-\frac{1}{2x}=-5\end{cases}}=>\orbr{\begin{cases}x\in\varnothing\\2x=\frac{1}{5}\end{cases}}=>x=\frac{1}{5}:2=>x=\frac{1}{10}\)
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