Câu hỏi của Nguyễn Trung Sơn

Question

Mọi người ơi giúp em làm bao bài cũng được ah. undefined

Nguyễn Lê Phước Thịnh · Accepted Answer

Bài 5:

a: $\left(x-1\right)^3=27$

=>$\left(x-1\right)^3=3^3$

=>x-1=3

=>x=4

b: $x^2+x=0$

=>x(x+1)=0

=>$\left[\begin{array}{l}x=0\ x+1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\ x=-1\end{array}\right.$

c: $\left(2x+1\right)^2=25$

=>$\left[\begin{array}{l}2x+1=5\ 2x+1=-5\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=4\ 2x=-6\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\ x=-3\end{array}\right.$

d: $\left(2x-3\right)^2=36$

=>$\left[\begin{array}{l}2x-3=6\ 2x-3=-6\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=9\ 2x=-3\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac92\ x=-\frac32\end{array}\right.$

e: $\left(2x-1\right)^3=-8$

=>2x-1=-2

=>2x=-2+1=-1

=>$x=-\frac12$

f: $5^{x+2}=625$

=>$5^{x+2}=5^4$

=>x+2=4

=>x=2

g: $\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}$

=>$\left(x-1\right)^{x+4}-\left(x-1\right)^{x+2}=0$

=>$\left(x-1\right)^{x+2}\cdot\left\lbrack\left(x-1\right)^2-1\right\rbrack=0$

=>$x\left(x-1\right)^{x+2}\cdot\left(x-2\right)=0$

=>x∈{0;1;2}

Bài 3:

a: $\left(0,25\right)^3\cdot32=\left(\frac14\right)^3\cdot32=32\cdot\frac{1}{64}=\frac12$

b: $\left(-0,125\right)^3\cdot80^4=\left(-0,125\cdot80\right)^3\cdot80=\left(-10\right)^3\cdot80=-1000\cdot80=-80000$

c: $\frac{8^2\cdot4^5}{2^{20}}=\frac{2^6\cdot2^{10}}{2^{20}}=\frac{1}{2^4}=\frac{1}{16}$

d: $\frac{81^{11}\cdot3^{17}}{27^{10}\cdot9^{15}}=\frac{3^{44}\cdot3^{17}}{3^{30}\cdot3^{30}}=\frac{3^{61}}{3^{60}}=3$

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