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\(\left(2x-3\right)^2=25\)
\(\left(2x-3\right)^2=5^2\)
\(\Rightarrow2x-3=5\)
\(\Rightarrow2x=8\)
\(\Rightarrow x=4\)
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a, (2x-3)3 = -64
=> (2x-3)3 = -43
=> 2x-3=-4
=> 2x = -1
=> x = -1 : 2
=> x = -1/2
b, (2x-3)2 =25
=> (2x-3)2 =5^2
=> 2x-3 = 5
=> 2x = 8
=> x = 4
c, (3x-4)2 =36
=> (3x-4)2 =62
=> 3x-4 = 6
=> 3x = 10
=> x = 3.(3)
d, 2x+1 = 64
=> 2x+1 = 26
=> x+1 = 6
=> x = 5
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a)(2x+1)2=25=52=(-5)2
=>2x+1=5 hoặc 2x+1=-5
2x=5-1 2x=-5-1
2x=4 2x=-6
x=4/2 x=-6/2
x=2 x=-3
Vậy x=2 hoặc x=-3
b)(2x-3)2=36=62=(-6)2
=>2x-3=6 hoặc 2x-3=-6
2x=6+3 2x=-6+3
2x=9 2x=-3
x=9/2 x=-3/2
Vậy x=9/2 hoặc x=-3/2
(2x+1)2 = 25
=> 2x2 + 12 = 25
=> 2x2 + 1 = 25
=> 2x2 = 24
=> x2 = 12
=> x = \(\sqrt{12}\)
(2x - 3)2 = 36
=> 2x2 - (32) = 36
=> 2x2 - 9 = 36
=> 2x2 = 45
=> x2 = 22,5
=> x = \(\sqrt{22,5}\)
t i c k nha!! 65768769789890
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a, \(2x\left(x-5\right)-x\left(2x+3\right)=25\)
\(\Rightarrow2x^2-10x-2x^2-3x=25\)
\(\Rightarrow-13x=25\Rightarrow x=\dfrac{-25}{13}\)
b, \(\left(3y^2-y+1\right)\left(y-1\right)+y^2\left(4-3y\right)=\dfrac{5}{2}\)
\(\Rightarrow3y^3-y^2+y-3y^2+y-1+4y^2-3y^3=\dfrac{5}{2}\)
\(\Rightarrow2y-1=\dfrac{5}{2}\Rightarrow2y=\dfrac{7}{2}\Rightarrow y=\dfrac{7}{4}\)
c, \(2x^2+3\left(x-1\right)\left(x+1\right)=5x\left(x+1\right)\)
\(\Rightarrow2x^2+3\left(x^2+x-x-1\right)=5x^2+5x\)
\(\Rightarrow2x^2+3x^2-3=5x^2+5x\)
\(\Rightarrow-5x=3\Rightarrow x=\dfrac{-3}{5}\)
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\(\left(2x+3\right)^2=25\)
\(\Leftrightarrow\left(2x+3\right)^2=5^2=\left(-5\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}2x+3=5\\2x+3=-5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-4\end{cases}}\)
Vậy ...
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a) 2x.3/5 - 5x = 3/2 - 7x
6x/5 - 5x = 3/2 - 7x
-19x/5 = 3/2 - 7x
-19x = 15/2 - 35x
-19x + 35x = 15/2
16x = 15/2
x = 15/2 : 16
x = 15/32
b) (x - 1/5)2 = 4/25
(x - 1/5)2 = (+-2/5)2
x - 1/5 = +-2/5
x - 1/5 = 2/5 hoặc x - 1/5 = -2/5
x = 2/5 + 1/5 x = -2/5 + 1/5
x = 3/5 x = -1/5
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Mấy bạn làm thiếu rồi
Ta có
\(\left(2x+3\right)^2=25=5^2=\left(-5\right)^2\)
Suy ra \(\left\{{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\)
Suy ra \(\left\{{}\begin{matrix}x=\dfrac{5-3}{2}=1\\x=\dfrac{-5-3}{2}=-4\end{matrix}\right.\)
Vậy x=1 và x=-4
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\(a,\left(x-3\right)^2=1\)
=> \(\sqrt{\left(x-3\right)^2}=\sqrt{1}\)
=> \(\left|x-3\right|=1\)
=> \(\left[{}\begin{matrix}x-3=1\\x-3=-1\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=1+3=4\\x=-1+3=2\end{matrix}\right.\)
Vậy \(x\in\left\{4;2\right\}\)
\(\left(2x+1\right)^3=-8\)
=> \(\sqrt[3]{\left(2x+1\right)^3}=\sqrt[3]{-8}\)
=> \(2x+1=-2\)
=> \(2x=-2-1=-3\)
=> \(x=-3:2=-\frac{3}{2}\)
Vậy \(x\in\left\{-\frac{3}{2}\right\}\)
\(c,\left(x-\frac{1}{4}\right)^2=\frac{1}{25}\)
=> \(\sqrt{\left(x-\frac{1}{4}\right)^2}=\sqrt{\frac{1}{25}}\)
=> \(\left|x-\frac{1}{4}\right|=\frac{1}{5}\)
=> \(\left[{}\begin{matrix}x-\frac{1}{4}=\frac{1}{5}\\x-\frac{1}{4}=-\frac{1}{5}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\frac{1}{5}+\frac{1}{4}=\frac{9}{20}\\x=-\frac{1}{5}+\frac{1}{4}=\frac{1}{20}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{9}{20};\frac{1}{20}\right\}\)
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\(a,\left(\dfrac{2}{3}\right)^x=\left(\dfrac{4}{9}\right)^4.\)
\(\left(\dfrac{2}{3}\right)^x=\left[\left(\dfrac{2}{3}\right)^2\right]^4.\)
\(\left(\dfrac{2}{3}\right)^x=\left(\dfrac{2}{3}\right)^8\Rightarrow x=8.\)
Vậy.....
\(b,\left(2x-1\right)^2=25.\)
\(\left(2x-1\right)^2=\left(\pm5\right)^2.\)
\(\Rightarrow\left(2x-1\right)=\pm5.\)
+) Xét \(2x-1=5\), ta có:
\(2x-1=5.\)
\(\Rightarrow2x=6.\)
\(\Rightarrow x=3.\)
+) Xét \(2x-1=-5\), ta có:
\(2x-1=-5.\)
\(\Rightarrow2x=-4.\)
\(\Rightarrow x=-2.\)
Vậy.....
\(\Leftrightarrow\left(2x+3\right)^2=5^2\Rightarrow2x+3=\pm5\)
+ Với \(2x+3=5\Rightarrow x=1\)
+ Với \(2x+3=-5\Rightarrow x=-4\)