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a,Ta có: \(\left(x+2\right)^4=\left(x+2\right)^6\)
\(\left(x+2\right)^4-\left(x+2\right)^6=0\)
\(\left(x+2\right)^4\text{[}1-\left(x+2\right)^2\text{]=0}\)
\(\Rightarrow\orbr{\begin{cases}\left(x+2\right)^4=0\\1-\left(x+2\right)^2=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\x+2=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-1\end{cases}}\)
Ta có : 2x + 2x + 1 = 24
=> 2x(1 + 2) = 24
=> 2x.3 = 24
=> 2x = 8
=> 2x = 23
=> x = 3
Ta có : (x + 2)4 = (x + 2)6
=> (x + 2)4 - (x + 2)6 = 0
<=> (x + 2)4 (1 - (x + 2)2) = 0
<=> \(\orbr{\begin{cases}\left(x+2\right)^4=0\\\left(1-\left(x+2\right)^2\right)=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x+2=0\\\left(x+2\right)^2=1\end{cases}}\)
<=> \(\orbr{\begin{cases}x+2=0\\x+2=1\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-2\\x=-1\end{cases}}\)
2x + 2x+1 = 24
=> 2x + 2x+1 = 23 + 24
2x+1 = 24
x+1 = 4
x = 4 - 1
x = 3
Vậy x lần lượt là 3 và 4
\(\Leftrightarrow\)2x + 2x . 2 = 24
\(\Leftrightarrow\)2x . (1 + 2 ) = 24
\(\Leftrightarrow\)2x . 3 = 24
\(\Leftrightarrow\)2x = 24 : 3
\(\Leftrightarrow\)2x = 8
\(\Leftrightarrow\)2x = 23
\(\Leftrightarrow\)x = 3
Ta thấy : \(\left(x-y^2+z\right)^2\ge0\forall x,y,z\)
\(\left(y-2\right)^2\ge0\forall y\)
\(\left(z+3\right)^2\ge0\forall z\)
Do đó : \(\left(x-y^2+z\right)^2+\left(y-2\right)^2+\left(z+3\right)^2\ge0\forall x,y,z\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x-y^2+z\right)^2=0\\\left(y-2\right)^2=0\\\left(z+3\right)^2=0\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x-y^2+z=0\\y-2=0\\z+3=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x-2^2+\left(-3\right)=0\\y=2\\z=-3\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x=7\\y=2\\z=-3\end{cases}}\)
Vậy : \(\left(x,y,z\right)=\left(7,2,-3\right)\)
5.(x-4)=123-38
5.(x-4)=85
x-4=85:5
x-4=17
x=17+4
x=21
Đúng 1000000000000000
bài cuối đây:
(x+1)+(x+2)+(x+3)+...+(x+100)=5750
[(x+100)+(x+1)].100 /2 =5750
(2x+101).100 /2 =5750
(2x+101).50=5750
2x+101=115
2x=14
x=7
(X + 2)4 = (x + 2)6
(x + 2)6 - (x + 2)4 = 0
(x + 2)4.[(x + 2)2 - 1] = 0
\(\Rightarrow\orbr{\begin{cases}\left(x+2\right)^4=0\\\left(x+2\right)^2-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x+2=0\\\left(x+2\right)^2=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=-2\\\end{cases}}\)
\(\left(x+2\right)^2=1\Rightarrow\orbr{\begin{cases}x+2=-1\\x+2=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\x=1\end{cases}}\)
=> x = {-3 ; -2 ; 1}