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a. (x-1/20)2=0
=> x-1/20=0
=> x=1/20
b. (x-2)2=1
=> (x-2)2=12=(-1)2
+) x-2=1
=> x=3
+) x-2=-1
=> x=1
Vậy x \(\in\){1;3}
c. (2x-1)3=-8
=> (2x-1)3=(-2)3
=> 2x-1=-2
=> 2x=-1
=> x=-1/2
d. (x+1/2)2=1/16
=> (x+1/2)2=(1/4)2=(-1/4)2
+) x+1/2=1/4
=> x=-1/4
+) x+1/2=-1/4
=> x=-3/4
Vậy x \(\in\){-3/4; -1/4}
a) \(\left(x-\frac{1}{2}\right)^2=0\Rightarrow x-\frac{1}{2}=0\Rightarrow x=\frac{1}{2}\)
b) \(\left(x-2\right)^2=1\)
\(\Rightarrow\orbr{\begin{cases}x-2=1\\x-2=-1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}\)
c)\(\left(2x-1\right)^3=-8=\left(-2\right)^3\)
\(\Rightarrow2x-1=-2\)
\(2x=-1\)
\(x=-\frac{1}{2}\)
d) \(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)
\(\Rightarrow\orbr{\begin{cases}x+\frac{1}{2}=\frac{1}{4}\\x+\frac{1}{2}=-\frac{1}{4}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-\frac{1}{4}\\x=-\frac{3}{4}\end{cases}}\)
1) Các cách viết số 25 dưới dãng lũy thừa là: 251; 52; (-5)2
2) a) \(\left(x-\frac{1}{2}\right)^2=0\)
=> \(x-\frac{1}{2}=0\)
=> \(x=\frac{1}{2}\)
Vậy \(x=\frac{1}{2}\)
b) (x - 2)2 = 1
=> \(\left[\begin{array}{nghiempt}x-2=1\\x-2=-1\end{array}\right.\)=> \(\left[\begin{array}{nghiempt}x=3\\x=1\end{array}\right.\)
Vậy \(x\in\left\{3;1\right\}\)
c) (2x - 1)3 = -8
=> (2x - 1)3 = (-2)3
=> 2x - 1 = -2
=> 2x = -2 + 1
=> 2x = -1
=> \(x=-\frac{1}{2}\)
Vậy \(x=-\frac{1}{2}\)
d) \(\left(x+\frac{1}{2}\right)^2=16\)
=> \(\left[\begin{array}{nghiempt}x+\frac{1}{2}=\frac{1}{4}\\x+\frac{1}{2}=-\frac{1}{4}\end{array}\right.\)=> \(\left[\begin{array}{nghiempt}x=-\frac{1}{4}\\x=-\frac{3}{4}\end{array}\right.\)
Vậy \(x\in\left\{-\frac{1}{4};-\frac{3}{4}\right\}\)
1) Các cách viết số 25 dưới dãng lũy thừa là: 251; 52; (-5)2
2) a) (x−12)2=0(x−12)2=0
=> x−12=0x−12=0
=> x=12x=12
Vậy x=12x=12
b) (x - 2)2 = 1
=> [x−2=1x−2=−1[x−2=1x−2=−1=> [x=3x=1[x=3x=1
Vậy x∈{3;1}x∈{3;1}
c) (2x - 1)3 = -8
=> (2x - 1)3 = (-2)3
=> 2x - 1 = -2
=> 2x = -2 + 1
=> 2x = -1
=> x=−12x=−12
Vậy x=−12x=−12
d) (x+12)2=16(x+12)2=16
=> [x+12=14x+12=−14[x+12=14x+12=−14=> [x=−14x=−34[x=−14x=−34
Vậy x∈{−14;−34}
`#3107.101107`
`1.`
`a,`
`(2x - 3)^2 = |3 - 2x|`
`=> (2x - 3)^2 = |2x - 3|`
`=>`\(\left[{}\begin{matrix}2x-3=\left(2x-3\right)^2\\2x-3=-\left(2x-3\right)^2\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}2x-3-\left(2x-3\right)^2=0\\2x-3+\left(2x-3\right)^2=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}\left(2x-3\right)\left(1-2x+3\right)=0\\\left(2x-3\right)\left(1+2x-3\right)=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}2x-3=0\\4-2x=0\\2x-2=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=2\\x=1\end{matrix}\right.\)
Vậy, `x \in {3/2; 2; 1}`
`b,`
`(x - 1)^2 + (2x - 1)^2 = 0`
`=>`\(\left[{}\begin{matrix}\left(x-1\right)^2=0\\\left(2x-1\right)^2=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x-1=0\\2x-1=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=1\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy, `x \in {1; 1/2}`
`c,`
`5 - x^2 = 1`
`=> x^2 = 4`
`=> x^2 = (+-2)^2`
`=> x = +-2`
Vậy, `x \in {-2; 2}`
`d,`
`x - 2\sqrt{x} = 0`
`=> x^2 - (2\sqrt{x})^2 = 0`
`=> x^2 - 4x = 0`
`=> x(x - 4) = 0`
`=>`\(\left[{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
Vậy, `x \in {0; 4}`
`g,`
`(x - 1) + 1/7 = 0`
`=> x - 1 + 1/7 = 0`
`=> x - 6/7 = 0`
`=> x = 6/7`
Vậy, `x = 6/7.`
Bài 1:
a) Ta có: \(\dfrac{17}{6}-x\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)
\(\Leftrightarrow\dfrac{17}{6}-x^2+\dfrac{7}{6}x-\dfrac{7}{4}=0\)
\(\Leftrightarrow-x^2+\dfrac{7}{6}x+\dfrac{13}{12}=0\)
\(\Leftrightarrow-12x^2+14x+13=0\)
\(\Delta=14^2-4\cdot\left(-12\right)\cdot13=196+624=820\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{14-2\sqrt{205}}{-24}=\dfrac{-7+\sqrt{205}}{12}\\x_2=\dfrac{14+2\sqrt{2015}}{-24}=\dfrac{-7-\sqrt{205}}{12}\end{matrix}\right.\)
b) Ta có: \(\dfrac{3}{35}-\left(\dfrac{3}{5}-x\right)=\dfrac{2}{7}\)
\(\Leftrightarrow\dfrac{3}{5}-x=\dfrac{3}{35}-\dfrac{10}{35}=\dfrac{-7}{35}=\dfrac{-1}{5}\)
hay \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)
a) \(\left(x-\frac{1}{2}\right)^2=0\)
\(x-\frac{1}{2}=0\)
\(x=0+\frac{1}{2}\)
\(x=\frac{1}{2}\)
b) \(\left(x-2\right)^2=1\)
\(\left(x-2\right)^2=1^2\)
\(x-2=1\)
\(x=1+2\)
\(x=3\)
c) \(\left(2x-1\right)^3=\left(-8\right)\)
\(\left(2x-1\right)^3=\left(-2\right)^3\)
\(2x-1=\left(-2\right)\)
\(2x=\left(-2\right)+1\)
\(2x=-1\)
\(x=-\frac{1}{2}\)
d) \(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)
\(\left(x+\frac{1}{2}\right)^2=\left(\frac{1}{4}\right)^2\)
\(x+\frac{1}{2}=\frac{1}{4}\)
\(x=\frac{1}{4}-\frac{1}{2}\)
\(x=-\frac{1}{4}\)
a) \(\left(x-\frac{1}{2}\right)^2=0\)
\(\Leftrightarrow x-\frac{1}{2}=0\)
\(\Leftrightarrow x=\frac{1}{2}\)
b) \(\left(x-2\right)^2=1\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=1\\x-2=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}}\)
c) \(\left(2x-1\right)^2=-8\)
\(\Leftrightarrow2x-1=-2\)
\(\Leftrightarrow2x=-1\)
\(\Leftrightarrow x=-\frac{1}{2}\)
d) \(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)
\(\Rightarrow\orbr{\begin{cases}x+\frac{1}{2}=\frac{1}{4}\\x+\frac{1}{2}=-\frac{1}{4}\end{cases}\Rightarrow\orbr{\begin{cases}x=-\frac{1}{4}\\x=-\frac{3}{4}\end{cases}}}\)
a) \(\left(x-\frac{1}{2}\right)^2=0\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=\frac{1}{2}\end{cases}}\)
b) \(\left(x-2\right)^2=1\)
\(\Leftrightarrow\left(x-2\right)^2-1=0\)
\(\Leftrightarrow\left(x-2-1\right)\left(x-2+1\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}}\)
c) \(\left(2x-1\right)^3=-8\)
\(\Leftrightarrow2x-1=-2\)
\(\Leftrightarrow2x=-1\)
\(\Leftrightarrow x=\frac{-1}{2}\)
d) \(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)
\(\Rightarrow x+\frac{1}{2}=\frac{1}{4}\)
\(\Rightarrow x=-\frac{1}{4}\)
a.
\(\left(x-\frac{1}{2}\right)^2=0\)
\(x-\frac{1}{2}=0\)
\(x=\frac{1}{2}\)
b.
\(\left(x-2\right)^2=1\)
\(x-2=\pm1\)
TH1:
\(x-2=1\)
\(x=1+2\)
\(x=3\)
TH2:
\(x-2=-1\)
\(x=-1+2\)
\(x=1\)
Vậy x = 3 hoặc x = 1
c.
\(\left(2x-1\right)^3=-8\)
\(\left(2x-1\right)^3=\left(-2\right)^3\)
\(2x-1=-2\)
\(2x=-2+1\)
\(2x=-1\)
\(x=-\frac{1}{2}\)
d.
\(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)
\(\left(x+\frac{1}{2}\right)^2=\left(\pm\frac{1}{4}\right)^2\)
\(x+\frac{1}{2}=\pm\frac{1}{4}\)
TH1:
\(x+\frac{1}{2}=\frac{1}{4}\)
\(x=\frac{1}{4}-\frac{1}{2}\)
\(x=\frac{1}{4}-\frac{2}{4}\)
\(x=-\frac{1}{4}\)
TH2:
\(x+\frac{1}{2}=-\frac{1}{4}\)
\(x=-\frac{1}{4}-\frac{1}{2}\)
\(x=-\frac{1}{4}-\frac{2}{4}\)
\(x=-\frac{3}{4}\)
Vậy \(x=-\frac{1}{4}\) hoặc \(x=-\frac{3}{4}\)
a) (x-1/2)2 = 0
=>x-1/2=0
x=1/2
b) (x-2)2 = 1
=>x-2=1 hoặc x-2=-1
x=3 hoặc x=1
c) (2x-1)3 = 8
(2x-1)3=23
=>2x-1=3
2x=4
x=2
d) (x+1/2)2 = 1/16
(x+1/2)2=(1/4)2
=>x+1/2=1/4
x=1/4-1/2
x=1/4-2/4
x=-1/4
{x-2020}^20-2019-1=0
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