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a)
\(\left(2x-3\right)^2=36\)
\(\Rightarrow\left[\begin{array}{nghiempt}\left(2x-3\right)^2=6^2\\\left(2x-3\right)^2=\left(-6\right)^2\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}2x-3=6\\2x-3=-6\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{9}{2}\\x=-\frac{3}{2}\end{array}\right.\)
b)
\(\left(x+2\right)^4=16\)
\(\Rightarrow\left[\begin{array}{nghiempt}\left(x+2\right)^4=2^4\\\left(x+2\right)^4=\left(-2\right)^4\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x+2=2\\x+2=-2\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x=-2\end{array}\right.\)
c)
\(\left(3x+2\right)^3=81\)
\(\Rightarrow\left(3x+2\right)^3=3^3\)
\(\Rightarrow3x+2=3\)
\(\Rightarrow x=\frac{1}{3}\)
a) \(\left(2x-3\right)^2=36\)
\(\Leftrightarrow\left(2x-3\right)^2=6^2=\left(-6\right)^2\)
\(\Leftrightarrow\begin{cases}2x-3=6\\2x-3=-6\end{cases}\)\(\Leftrightarrow\begin{cases}2x=9\\2x=-3\end{cases}\)\(\Leftrightarrow\begin{cases}x=4.5\\x=-1.5\end{cases}\)
Vậy.....
b) \(\left(x+2\right)^4=16\)
\(\Leftrightarrow\left(x+2\right)^4=2^4=\left(-2\right)^4\)
\(\Leftrightarrow\begin{cases}x+2=2\\x+2=-2\end{cases}\)\(\Leftrightarrow\begin{cases}x=0\\x=-4\end{cases}\)
Vậy........
Ta có : \(\sqrt{x}\) = 3 ⇒ x = \(3^2\)= 9
⇒ \(x^2\) = \(9^2\) = 81
Vậy ta chọn D
\(a)\dfrac{y+z+1}{x}=\dfrac{z+x+2}{y}=\dfrac{x+y-3}{z}=\dfrac{1}{x+y+z}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}=\dfrac{y+z+x+x+z+2+x+y-3}{x+y+z}\)
\(=\dfrac{\left(x+y+z\right)+\left(x+y+z\right)+\left(1+2-3\right)}{x+y+z}=\dfrac{2\left(x+y+z\right)}{x+y+z}=2\)
Lại có: \(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}=\dfrac{1}{x+y+z}\)
\(\Rightarrow2=\dfrac{1}{x+y+z}\Rightarrow2\left(x+y+z\right)=1\Rightarrow x+y+z=\dfrac{1}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{y+z+1}{x}=2\\\dfrac{x+z+2}{y}=2\\\dfrac{x+y-3}{z}=2\\x+y+z=\dfrac{1}{2}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y+z+1=2x\\x+z+2=2y\\x+y-3=2z\\x+y+z=\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}y+z+x+1=3x\\x+y+z+2=3y\\x+y+z-3=3z\\x+y+z=\dfrac{1}{2}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{2}+1=3x\\\dfrac{1}{2}+2=3y\\\dfrac{1}{2}-3=3z\\x+y+z=\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1+\dfrac{1}{2}}{3}\\y=\dfrac{\dfrac{1}{2}+2}{3}\\z=\dfrac{\dfrac{1}{2}-3}{3}\\x+y+z=\dfrac{1}{2}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{5}{6}\\z=\dfrac{-5}{6}\end{matrix}\right.\)
Chúc bạn học tốt!
a) x^2 +xy+ax+ay
=x(x+y)+a(x+y)
=(x+y)(x+a)
mã đề có phải là phân tích thành nhân tử ko bạn