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a)
\(\left(\frac{x-1}{4}\right)^2=\frac{4}{9}\)
⇒ \(\frac{x-1}{4}=\frac{2}{3}\)
\(\Rightarrow\frac{\left(x-1\right).3}{12}=\frac{8}{12}\)
=> (x - 1) = 8/3
=> x = 8/3 - 1
=> x = 5/3
a)\(\frac{9}{4}\cdot\left|x\right|-\frac{5}{2}=\frac{8}{3}\)\(\Rightarrow\frac{9}{4}\cdot\left|x\right|=\frac{8}{3}+\frac{5}{2}\Rightarrow\frac{9}{4}\cdot\left|x\right|=\frac{31}{6}\)
\(\Rightarrow\left|x\right|=\frac{31}{6}:\frac{9}{4}\Rightarrow\left|x\right|=\frac{31}{6}\cdot\frac{4}{9}\Rightarrow\left|x\right|=\frac{62}{27}\)
\(\Rightarrow\hept{\begin{cases}x=\frac{62}{27}\\x=-\frac{62}{27}\end{cases}}\)
b)\(\frac{1}{2}\cdot\left|x\right|+\frac{3}{4}=\frac{2}{3}\Rightarrow\frac{1}{2}\cdot\left|x\right|=\frac{2}{3}-\frac{3}{4}\Rightarrow\frac{1}{2}\cdot\left|x\right|=-\frac{1}{12}\)
\(\Rightarrow\left|x\right|=-\frac{1}{12}:\frac{1}{2}\Rightarrow\left|x\right|=-\frac{1}{12}\cdot2\Rightarrow\left|x\right|=-\frac{1}{6}\)
Ta có\(\left|x\right|\ge0\)mà \(-\frac{1}{6}\le0\)
Do đó ko có giá trị của x thỏa mãn
a) 1/4(x-3)+2=1/5
1/4.(x-3) = 1/5-2
1/4.(x-3) = -9/5
x-3 = (-9/5):1/4
x-3 = -36/5
x = -36/5+3
x= -21/5
1) \(5-\left(1+\dfrac{1}{3}\right):\left(1-\dfrac{1}{3}\right)\)
\(=5-\dfrac{4}{3}:\dfrac{2}{3}\)
\(=5-\dfrac{4}{3}\cdot\dfrac{3}{2}\)
\(=5-\dfrac{4}{2}\)
\(=5-2\)
\(=3\)
b) \(\left(1+\dfrac{2}{3}-\dfrac{5}{4}\right)-\left(1-\dfrac{5}{4}\right)+2022-\dfrac{2}{3}\)
\(=1+\dfrac{2}{3}-\dfrac{5}{4}-1+\dfrac{5}{4}++2022-\dfrac{2}{3}\)
\(=\left(1-1\right)+\left(\dfrac{2}{3}-\dfrac{2}{3}\right)+\left(-\dfrac{5}{4}+\dfrac{5}{4}\right)+2022\)
\(=0+0+0+2022\)
\(=2022\)
2) \(0,7^2\cdot x=0,49^2\)
\(\Rightarrow x=\dfrac{0,49^2}{0,7^2}\)
\(\Rightarrow x=\left(\dfrac{0,49}{0,7}\right)^2\)
\(\Rightarrow x=\left(0,7\right)^2\)
\(\Rightarrow x=0,49\)
b) \(x:\left(-0,5\right)^3=\left(0,5\right)^2\)
\(\Rightarrow x=\left(0,5\right)^2\cdot\left(-0,5\right)^3\)
\(\Rightarrow x=\left(-0,5\right)^5\)
\(\Rightarrow x=-\dfrac{1}{32}\)
2:
a: =>x*0,49=0,49^2
=>x=0,49
b: =>x=(0,5)^2*(-1)*(0,5)^3=-(0,5)^5
a) (2x)5 : 43 = 815 => 25x = 815.43 = (23)15.(22)3 = 245.26 = 251 => 5x = 51 => x = 10,2
b) (32)x .93 = 2439 => 32x = 2439 : 93 = (35)9 : (32)3 = 345 : 36 = 339 => 2x = 39 => x = 19,5
c) (1/125)3.5x = 255 => 5x = 255 : (1/125)3 = (52)5 : (1/53)3 = 510 : (5-3)3 = 510 : 5-9 = 519 => x = 19
d) 1/81 : 3x = 1/729 => 3x = 1/81 : 1/729 = 1/34.729 = 3-4.36 = 32 => x = 2
e) (5x - 2)4 = 168 = (162)4 = 2564
=> 5x - 2 = -256 ; 256 => 5x = -254 ; 258 => x = -50,8 ; 51,6
P/S : Thay x = 10,2 vào câu a , x = 19,5 vào câu b sẽ thấy điều hư cấu : 210,2 và 919,5.Ko thể tính được giá trị của 2 lũy thừa này.
a, \(\left(x-\frac{1}{4}\right)^2=\frac{4}{9}=>\left(x-\frac{1}{4}\right)=\sqrt{\frac{4}{9}}=\frac{2}{3}\)
=> \(x=\frac{1}{4}+\frac{2}{3}=\frac{11}{12}\)
b, \(\left(x+0,7\right)^3=-8=>\left(x+0,7\right)^3-2^3\)
=> \(x+0,7=-2=>x=-2-0,7=-2,7\)
c \(2^x+2^{x+3}=144=>2^x+2^x.8=144\)
=>\(2^x.\left(8+1\right)=144=>2^x.9=144=>2^x=16\)
=> \(2^x=2^4=>x=4\)
d, \(5-2:\left|x-2\right|=1=>2:\left|x-1\right|=4\)
=> \(\left|x-2\right|=\frac{2}{4}=\frac{1}{2}\)
vậy \(x-2=\frac{1}{2}hoacx-2=-\frac{1}{2}\)
x = \(\frac{2}{5}\) x = \(\frac{3}{2}\)
a )
\(\left(x-\frac{1}{4}\right)^2=\frac{4}{9}\)
\(\left(x-\frac{1}{4}\right)^2=\left(\pm\frac{2}{3}\right)^2\)
\(\Rightarrow x-\frac{1}{4}=\frac{2}{3}\) hoặc \(x-\frac{1}{4}=-\frac{2}{3}\)
\(x=\frac{2}{3}+\frac{1}{4}\) hoặc \(x=-\frac{2}{3}+\frac{1}{4}\)
\(x=\frac{8}{12}+\frac{3}{12}\) hoặc \(x=-\frac{8}{12}+\frac{3}{12}\)
\(x=\frac{11}{12}\) hoặc \(x=-\frac{5}{12}\)
b)
\(\left(x+0,7\right)^3=-8\)
\(\left(x+0,7\right)^3=\left(-2\right)^3\)
\(\Rightarrow x+0,7=-2\)
\(x=-2-0,7\)
\(x=-2,7\)
c)
\(2^x+2^{x+3}=144\)
\(2^x\cdot1+2^x\cdot2^3=144\)
\(2^x\cdot\left(1+2^3\right)=144\)
\(2^x\cdot\left(1+8\right)=144\)
\(2^x\cdot9=144\)
\(2^x=144:9\)
\(2^x=16\)
\(2^x=2^4\) \(\Rightarrow x=4\)
d)
\(5-2:\left|x-2\right|=2\)
\(5-\left|x-2\right|=2\cdot2\)
\(5-\left|x-2\right|=4\)
\(\left|x-2\right|=5-4\)
\(\left|x-2\right|=1\)
\(\left|x-2\right|=\pm1\)
\(x-2=1\) hoặc \(x-2=-1\)
\(x=1+2\) hoặc \(x=-1+2\)
\(x=3\) hoặc \(x=1\)