\(\left(x-\frac{1}{3}\right)\left(y-\frac{1}{2}\right)\left(z-5\right)=0\)

\(\Rightarrow\hept{\begin{cases}x=\frac{1}{3}\\y=\frac{1}{2}\\z=5\end{cases}}\)

Vì \(z+3=y+1\Rightarrow y=7\)

Lại có \(y+1=x+2\Rightarrow x=8-2=6\)

Vậy x = 6 ; y = 7 ; z = 5

x=\(\frac{1}{3}\)

Chẳng hiểu gì hết trơn

a) \(\left(x-1\right)^2=0\Leftrightarrow x-1=0\Leftrightarrow x=1\) vậy \(x=1\)

b) \(\left(x-2\right)^2-1=0\Leftrightarrow\left(x-2\right)^2=1\) \(\Leftrightarrow\left\{{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=1\end{matrix}\right.\) vậy \(x=3;x=1\)

c) \(\left(2x-1\right)^3=-8\Leftrightarrow2x-1=\sqrt[3]{-8}\Leftrightarrow2x-1=-2\)

\(\Leftrightarrow2x=-1\Leftrightarrow x=\dfrac{-1}{2}\) vậy \(x=\dfrac{-1}{2}\)

d) \(\left(x+2\right)^2+1=0\Leftrightarrow\left(x+2\right)^2=-1\) (vô lí)

vậy phương trình vô nghiệm

a) (x-1)² = 0

<=> x-1 = 0

<=> x = 1

b) (x-2)² - 1 = 0

<=> (x-2)² = 1

<=> \(\left\{{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

c) (2x-1)³ = -8

<=> (2x-1)³ = -2³

<=> 2x - 1 = -2

<=> 2x = -1

<=> x = \(-\dfrac{1}{2}\)

d) (x+2)² + 1 = 0

<=> (x+2)² = -1

<=> x+2 = -1

<=> x = -3

a, \(\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(x-1\right)^2=0^2\)

\(\Leftrightarrow x-1=0\)

\(\Leftrightarrow x=1\)

Vậy ......

b, \(\left(x-2\right)^2-1=0\)

\(\Leftrightarrow\left(x-2\right)^2=1\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-2\right)^2=1^2\\\left(x-2\right)^2=\left(-1\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

Vậy .....

c, \(\left(2x-1\right)^3=-8\)

\(\Leftrightarrow\left(2x-1\right)^3=\left(-2\right)^3\)

\(\Leftrightarrow2x-1=-2\)

\(\Leftrightarrow2x=-3\)

\(\Leftrightarrow x=-\dfrac{3}{2}\)

Vậy ...

d, \(\left(x+2\right)^2+1=0\)

\(\Leftrightarrow\left(x+2\right)^2=-1\)

\(\Leftrightarrow\) ko tìm dc giá trị của x thỏa mãn (do \(\left(x+2\right)^2\ge0\))

=> (x-5)^x+1-(x-5)^x+1+20=0

=> (x-5)^x+1.1 - (x-5)^x+1.(x-5)²⁰=0

=> (x-5)^x+1.[1-(x-5)²⁰ ]=0

=> (x-5)^x+1 = 0 hoặc 1-(x-5)²⁰=0

=> x-5 = 0       hoặc (x-5)²⁰ = 1

=> x=5            hoặc x-5=1

=> x=5            hoặc x=6.

a) thay \(x-y=\frac{3}{10}\)vào \(y\left(x-y\right)=\frac{-3}{50}\)ta có\(\frac{3}{10}y=\frac{-3}{50}\)=>\(y=\frac{-3}{50}:\frac{3}{10}=\frac{-1}{5}\)=>\(x-y=\frac{3}{10}\Rightarrow x=\frac{3}{10}+\frac{-1}{5}=\frac{1}{10}\)

hôm sau mik giải tip cho

nick mik co vip do