Question

1)tìm x, biết

a)giá trị tuyệt đối x-2=x

b)giá trị tuyệt đối x-3.4+giá trị tuyệt đối 2.6-x=6

2)tính

\(M=\frac{8^{10}+4^{10}}{8^4+4^{11}}\)

3)tìm x, biết

a) \(\left(x+5\right)^3=-64\)

b) \(\left(2x-3\right)^2=9\)

c) \(x^2+1=82\)

d) \(x^2+\frac{7}{4}=\frac{23}{4}\)

e) \(\left(2x+3\right)^2=25\)

Answer 1

3)

a)\(\left(x+5\right)^3=-64\\ \Leftrightarrow\left(x+5\right)^3=\left(-4\right)^3\\ \Leftrightarrow x+5=-4\\ \Leftrightarrow x=-9\)

Vậy x = -9

b)\(\left(2x-3\right)^2=9\\ \Leftrightarrow\left(2x-3\right)^2=\left(\pm3\right)^2\\ \Rightarrow2x-3\in\left\{3;-3\right\}\Rightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)

Vậy...

c)\(x^2+1=82\\ \Leftrightarrow x^2=81\\ \Leftrightarrow x^2=\left(\pm9\right)^2\\ \Rightarrow x\in\left\{9;-9\right\}\)

Vậy...

d)\(x^2+\frac{7}{4}=\frac{23}{4}\\ \Leftrightarrow x^2=16\\ \Leftrightarrow x^2=\left(\pm4\right)^2\\ \Rightarrow x\in\left\{4;-4\right\}\)

Vậy...

e)\(\left(2x+3\right)^2=25\\ \Leftrightarrow\left(2x+3\right)^2=\left(\pm5\right)^2\\ \Rightarrow2x+3\in\left\{5;-5\right\}\\ \Rightarrow\left[{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)

Vậy...

Answer 2

3)

a) \(\left(x+5\right)^3=-64\)

\(\Rightarrow\left(x+5\right)^3=\left(-4\right)^3\)

\(\Rightarrow x+5=-4\)

\(\Rightarrow x=\left(-4\right)-5\)

\(\Rightarrow x=-9\)

Vậy \(x=-9.\)

b) \(\left(2x-3\right)^2=9\)

\(\Rightarrow2x-3=\pm3.\)

\(\Rightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=6\\2x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6:2\\x=0:2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)

Vậy \(x\in\left\{3;0\right\}.\)

c) \(x^2+1=82\)

\(\Rightarrow x^2=82-1\)

\(\Rightarrow x^2=81\)

\(\Rightarrow\left[{}\begin{matrix}x=9\\x=-9\end{matrix}\right.\)

Vậy \(x\in\left\{9;-9\right\}.\)

d) \(x^2+\frac{7}{4}=\frac{23}{4}\)

\(\Rightarrow x^2=\frac{23}{4}-\frac{7}{4}\)

\(\Rightarrow x^2=4\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

Vậy \(x\in\left\{2;-2\right\}.\)

Chúc bạn học tốt!

Answer 3

1.

a) \(\left|x-2\right|=x\)

\(\Rightarrow\left[{}\begin{matrix}x-2=x\\x-2=-x\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-x=2\\x+x=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}0=2\\2x=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x\in\varnothing\\x=1\left(TM\right)\end{matrix}\right.\)

Vậy \(x=1.\)

b) Đề là \(\left|x-3.4\right|+\left|2.6-x\right|=6\) à bạn?

Chúc bạn học tốt!