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\(a,\frac{2}{3}+\frac{7}{4}:x=\frac{5}{6}\)
\(\Leftrightarrow\frac{7}{4}:x=\frac{5}{6}-\frac{2}{3}\)
\(\Leftrightarrow\frac{7}{4}:x=\frac{1}{6}\)
\(\Leftrightarrow x=\frac{21}{2}\)
\(b,\left(x+\frac{5}{3}\right).\left(x-\frac{5}{4}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{5}{3}=0\\x-\frac{5}{4}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\frac{5}{3}\\\frac{5}{4}\end{matrix}\right.\)
\(c,\left(x-1,2\right)^2=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1,2=2\\x-1,2=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3,2\\x=-0,8\end{matrix}\right.\)
\(d,\left(x+1\right)^3=-125\)
\(\Leftrightarrow\left(x+1\right)^3=\left(-5\right)^3\)
\(\Leftrightarrow x+1=-5\)
\(\Leftrightarrow x=-6\)
Vậy ...........................................................
\(\left(x-1,2\right)^2=4\)
\(\left(x-1,2\right)^2=2^2\)
\(\Rightarrow\orbr{\begin{cases}x-1,2=2\\x-1,2=-2\end{cases}\Rightarrow\orbr{\begin{cases}x=3,2\\x=-0,8\end{cases}}}\)
\(\left(x+1\right)^3=-125\)
\(\left(x+1\right)^3=\left(-5\right)^3\)
\(x+1=-5\)
\(x=-6\)
\(a,\left(x-1,2\right)^2=4\)
\(x-1,2=\pm2\)
\(\orbr{\begin{cases}x-1,2=2\\x-1,2=-2\end{cases}\Rightarrow\orbr{\begin{cases}x=3,2\\x=-\frac{4}{5}\end{cases}}}\)
\(b,\left(x+1\right)^3=-125\)
\(x+1=-5\)
\(x=-6\)
a) \(2^x=8\)
⇔ \(2^x=2^3\)
⇒ \(x=3\)
b) \(3^x=27\)
⇔ \(3^x=3^3\)
⇒ \(x=3\)
c) \(\left(-\dfrac{1}{2}\right)x=\left(-\dfrac{1}{2}\right)^4\)
⇔ \(x=\left(-\dfrac{1}{2}\right)^4\div\left(-\dfrac{1}{2}\right)\)
⇔ \(x=\left(-\dfrac{1}{2}\right)^3\)
d) \(x\div\left(-\dfrac{3}{4}\right)=\left(-\dfrac{3}{4}\right)^2\)
⇔ \(x=\left(-\dfrac{3}{4}\right)^2\cdot\left(-\dfrac{3}{4}\right)\)
⇔ \(x=\left(-\dfrac{3}{4}\right)^3=-\dfrac{27}{64}\)
d) \(\left(x+1\right)^3=-125\)
⇔ \(\left(x+1\right)^3=\left(-5\right)^3\)
⇔ \(x+1=-5\)
⇔ \(x=-5-1=-6\)
2:
a: (x-1,2)^2=4
=>x-1,2=2 hoặc x-1,2=-2
=>x=3,2(loại) hoặc x=-0,8(loại)
b: (x-1,5)^2=9
=>x-1,5=3 hoặc x-1,5=-3
=>x=-1,5(loại) hoặc x=4,5(loại)
c: (x-2)^3=64
=>(x-2)^3=4^3
=>x-2=4
=>x=6(nhận)
\(\left(x-1,2\right)^2=4\)
⇔\(x^2-2.x.1,2+1,2^2=4\)
⇔\(x^2-2,4x+1,44=4\)
⇔\(x^2-2,4x=4-1,44\)
⇔\(x\left(x-2,4\right)=2,56\)
⇔\(x=2,56\) hoặc \(x-2,4=2,56\)
⇔\(x=2,56\) hoặc \(x=4,96\)
a) \(\left(x-1,2\right)^2=4=2^2\)
\(\Leftrightarrow x-1,2=4\)
\(\Leftrightarrow x=5,2\)
b) \(\left(x+1\right)^3=-125=\left(-5\right)^3\)
\(\Leftrightarrow x+1=-5\)
\(\Leftrightarrow x=-6\)
c) \(\left(x+1,5\right)^8+\left(2,7-y\right)^{10}=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+1,5=0\\2,7-y=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-1,5\\y=2,7\end{matrix}\right.\)
a) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\left(-\dfrac{7}{12}\right)\cdot1\dfrac{2}{5}\)
\(\Rightarrow\dfrac{1}{6}x=\left(-\dfrac{7}{12}\right)\cdot\dfrac{7}{5}\)
\(\Rightarrow\dfrac{1}{6}x=-\dfrac{49}{60}\)
\(\Rightarrow x=-\dfrac{49}{60}:\dfrac{1}{6}\)
\(\Rightarrow x=-\dfrac{49}{10}\)
b) \(\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2=\dfrac{9}{4}\)
\(\Rightarrow\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2=\left(\pm\dfrac{3}{2}\right)^2\)
+) \(\dfrac{1}{5}-\dfrac{3}{2}x=\dfrac{3}{2}\)
\(\Rightarrow\dfrac{3}{2}x=\dfrac{1}{5}-\dfrac{3}{2}\)
\(\Rightarrow\dfrac{3}{2}x=-\dfrac{13}{10}\)
\(\Rightarrow x=-\dfrac{13}{10}:\dfrac{3}{2}\)
\(\Rightarrow x=-\dfrac{13}{15}\)
+) \(\left(1,25-\dfrac{4}{5}x\right)^3=-125\)
\(\Rightarrow\left(\dfrac{5}{4}-\dfrac{4}{5}x\right)^3=\left(-5\right)^3\)
\(\Rightarrow\dfrac{5}{4}-\dfrac{4}{5}x=-5\)
\(\Rightarrow\dfrac{4}{5}x=\dfrac{5}{4}+5\)
\(\Rightarrow\dfrac{4}{5}x=\dfrac{25}{4}\)
\(\Rightarrow x=\dfrac{25}{4}:\dfrac{4}{5}\)
\(\Rightarrow x=\dfrac{125}{16}\)
a, \(\dfrac{2}{3}\)\(x\) - \(\dfrac{1}{2}\)\(x\) = (- \(\dfrac{7}{12}\)). 1\(\dfrac{2}{5}\)
\(x\).(\(\dfrac{2}{3}\) - \(\dfrac{1}{2}\)) = (- \(\dfrac{7}{12}\)) . \(\dfrac{7}{5}\)
\(x\). \(\dfrac{1}{6}\) = - \(\dfrac{49}{60}\)
\(x\) = - \(\dfrac{49}{60}\).6
\(x\) = -\(\dfrac{49}{10}\)
1.
a/ 1253 : 93 = \(\left(\frac{125}{9}\right)^3\) = \(\frac{1953125}{729}\)
b/ 324. 43 = \(\left(2^5\right)^4.\left(2^2\right)^3\) = \(2^{20}.2^6\) = \(2^{26}\) =67108864
2. \(\left(x+\frac{1}{2}\right)^2=\frac{4}{25}\)
=> \(\left(x+\frac{1}{2}\right)^{ }=\frac{2^2}{5^2}\)
=>\(\left(x+\frac{1}{2}\right)^2=\left(\frac{2}{5}\right)^2\)
=> \(x+\frac{1}{2}=\frac{2}{5}\)
=> \(x=\frac{2}{5}-\frac{1}{2}=\frac{-1}{10}\)
Vậy \(x=\frac{-1}{10}\)
a) /x/ = 2,5
=> x =2,5 hoặc x = -2,5
b)/x/ = -1,2
x ko là gì hết vì có dấu giá trị tuyệt đối mọ số sẽ thành dương
mà đây là số âm
c) /x/ + 0.573 = 2
=>/x/ = 2 - 0,573
=> /x/ = 1,427
=>x = 1,427 hoặc x = -1,427
d) /x+1/3/ -4 = -1
/x+1/3/=-1+4
/x+1/3/= 3
=>x +1/3 =3
hoặc x + 1/3 =-3
TH1 ; x + 1/3 = 3
x = 3 - 1/3
x = 8/3
TH2 ; x +1/3 = -3
x = -3 -1/3
x = -10/3
vậy x=8/3 hoặc x = -10/3
xin lỗi mik ko biết viết dấu giá trị tuyệt đôi
a) IxI = 2,5
Vậy x = 2,5 hoặc x = -2,5.
b) IxI = -1,2
=> Không có x thỏa mãn đề bài.
c) IxI + 0,573 = 2
Trường hợp 1:
x + 0,573 = 2
x = 2 - 0,573
x = 1,427
Trường hợp 2:
x + 0,573 = -2
x = -2 - 0,573
x = -2,573
Vậy x = {1,427 ; -2,573}
d) Ix + 1/3I - 4 = -1
Ix + 1/3I = -1+4
Ix + 1/3I = 3
Trường hợp 1:
x + 1/3 = 3
x = 3 - 1/3
x = 8/3
Trường hợp 2:
x + 1/3 = -3
x = -3 - 1/3
x = -10/3
a) |x| = 2,5
=>\(\left[\begin{array}{nghiempt}x=2,5\\x=-2,5\end{array}\right.\)
vậy x=2,5 hoặc x=-2,5
b)|x|=-1,2
=>x không có giá trị thỏa mãn |x|\(\ge\) 0
c)|x| + 0,573 = 2
|x| = 2 - 0,573
|x| = 1,427
=>\(\left[\begin{array}{nghiempt}x=1,427\\x=-1,427\end{array}\right.\)
Vậy x = 1,427 hoặc x = -1,427
d) ∣∣x+13∣∣ - 4 = -1
=>|x+\(\frac{1}{3}\)| =-1 + 4
|x+\(\frac{1}{3}\)| = 3
.....................
Vậy x = \(\frac{8}{3}\) hoặc x = \(\frac{-10}{3}\)
a ) \(\left|x\right|=2,5\Rightarrow x=2,5;x=-2,5\)
b ) \(\left|x\right|=-1,2\Rightarrow\left|x\right|\ge0\forall x\Rightarrow x\in\varnothing\)
c ) \(\left|x\right|+0,573=2\)
\(\Rightarrow\)\(\left[\begin{array}{nghiempt}x+0,573=2\\x+0,573=-2\end{array}\right.\) \(\Rightarrow\left[\begin{array}{nghiempt}2-0,573\\\left(-2\right)-0,573\end{array}\right.\) \(\Rightarrow\)\(\left[\begin{array}{nghiempt}x=1,427\\x=-2,573\end{array}\right.\)
Vậy \(x\in1,427;-2,573\)
d ) \(\left|x+\frac{1}{3}\right|-4=-1\)
\(\Rightarrow\left|x+\frac{1}{3}\right|=3\)
\(\Rightarrow\left[\begin{array}{nghiempt}x+\frac{1}{3}=3\\x+\frac{1}{3}=-3\end{array}\right.\) \(\Rightarrow\left[\begin{array}{nghiempt}x=3-\frac{1}{3}\\x=\left(-3\right)-\frac{1}{3}\end{array}\right.\) \(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{8}{3}\\x=\frac{-10}{3}\end{array}\right.\)
Vậy \(x\in\frac{8}{3};\frac{-10}{3}\)
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.
\(\left(a\right):\left(x-1,2\right)^2=4=\left(\pm2\right)^2\\ =>\left[{}\begin{matrix}x-1,2=2\\x-1,2=-2\end{matrix}\right.\\ =>\left[{}\begin{matrix}x=3,2\\x=-0,8\end{matrix}\right.\)
\(\left(b\right):\left(x+1\right)^3=-125=\left(-5\right)^3\\ =>x+1=-5\\ =>x=-6\)
chịu :)