Giúp mình với! Bài 2: Tìm x biết: ( Cùng số mũ),$1)~(x-1)^3=\frac18$ $2)~(3-x)^3=\frac{-27}{64}$ $3)~(x-5)^3=\frac1{-27}$,$4)~(x-4)^2=\frac49$ $5)~(5-x)^2=\frac1{16}$ $6)~(x+1)^2=\frac4{49}$

Question

Ninh Hoàng · Accepted Answer

{"userId":5421245,"userName":"THANH BÌNH"}

1)

$$\left(x-1\right)^3=\frac{1}{8}$$

$$\left(x-1\right)^3=\left(\frac{1}{2}\right)^3$$

$$x-1=\frac{1}{2}$$

$$x=\frac{3}{2}$$

2)

$$\left(3-x\right)^3=\frac{-27}{64}$$

$$\left(3-x\right)^3=\left(-\frac{3}{4}\right)^3$$

$$3-x=-\frac{3}{4}$$

$$x=\frac{15}{4}$$

3)

$$\left(x-5\right)^3=\frac{1}{-27}$$

$$\left(x-5\right)^3=\left(\frac{1}{-3}\right)^3$$

$$x-5=\frac{1}{-3}$$

$$x=\frac{14}{3}$$

4)

$$\left(x-4\right)^2=\frac{4}{9}$$

$$\left(x-4\right)^2=\left(\pm\frac{2}{3}\right)^2$$

$$x-4=\frac{2}{3}$$ hoặc $$x-4=-\frac{2}{3}$$

$$x=\frac{14}{3}$$ hoặc $$x=\frac{10}{3}$$

5)

$$\left(5-x\right)^2=\frac{1}{16}$$

$$\left(5-x\right)^2=\left(\pm\frac{1}{4}\right)^2$$

$$5-x=\frac{1}{4}$$ hoặc $$5-x=-\frac{1}{4}$$

$$x=\frac{19}{4}$$ hoặc $$x=\frac{21}{4}$$

6)

$$\left(x+1\right)^2=\frac{4}{49}$$

$$\left(x+1\right)^2=\left(\pm\frac{2}{7}\right)^2$$

$$x+1=\frac{2}{7}$$ hoặc $$x+1=-\frac{2}{7}$$

$$x=-\frac{5}{7}$$ hoặc $$x=-\frac{9}{7}$$.

Timi · Answer

Bài 2:
1) $(x-1)^3=\frac{1}{8}$
Ta có $\frac{1}{8}= (\frac{1}{2})^3$. 
Do đó $x-1=\frac{1}{2}$. 
Vậy $x=1+\frac{1}{2}=\frac{3}{2}$

2) $(3-x)^3=\frac{-27}{64}$
Ta có $\frac{-27}{64}= (\frac{-3}{4})^3$. 
Do đó $3-x=\frac{-3}{4}$. 
Vậy $x=3-\frac{-3}{4}=\frac{15}{4}$

3) $(x-5)^3=\frac{1}{-27}$
Ta có $\frac{1}{-27}= (\frac{-1}{3})^3$. 
Do đó $x-5=\frac{-1}{3}$. 
Vậy $x=5+(\frac{-1}{3})=\frac{14}{3}$

4) $(x-4)^2=\frac{4}{9}$
Ta có $\frac{4}{9}= (\frac{2}{3})^2$. 
Do đó $x-4=\frac{2}{3}$ hoặc $x-4=\frac{-2}{3}$. 
Vậy $x=4+\frac{2}{3}=\frac{14}{3}$ hoặc $x=4+(\frac{-2}{3})=\frac{10}{3}$

5) $(5-x)^2=\frac{1}{16}$
Ta có $\frac{1}{16}= (\frac{1}{4})^2$. 
Do đó $5-x=\frac{1}{4}$ hoặc $5-x=\frac{-1}{4}$. 
Vậy $x=5-(\frac{1}{4})=\frac{19}{4}$ hoặc $x=5-(\frac{-1}{4})=\frac{21}{4}$

6) $(x+1)^2=\frac{4}{49}$
Ta có $\frac{4}{49}= (\frac{2}{7})^2$. 
Do đó $x+1=\frac{2}{7}$ hoặc $x+1=\frac{-2}{7}$. 
Vậy $x=\frac{2}{7}-1=\frac{-5}{7}$ hoặc $x=\frac{-2}{7}-1=\frac{-9}{7}$