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Ở chỗ 6xx vs 2xx la 6x 2x nha cac ban
Cac ban lam on giai nhanh gium minh vs nhe
"X"="x" hay khác vậy
\(\left\{{}\begin{matrix}A=\dfrac{2x^2-6x+7}{x^2-2x+1}=\dfrac{2\left(x-1\right)^2-2\left(x-1\right)+3}{\left(x-1\right)^2}\\y=x-1\end{matrix}\right.\)
\(A=\dfrac{2y^2-2y+3}{y^2}=\dfrac{y^2-2.3y+9}{y^2}-\dfrac{5}{3}=\dfrac{\left(y-3\right)^2}{y^2}\ge\dfrac{5}{3}\)
\(A\ge\dfrac{5}{3}\) khi y=3=> x=4
b) \(x^2+2\sqrt{3}x-6=0\)
\(\Leftrightarrow\) \(x^2+2\sqrt{3}x+3-9=0\)
\(\Leftrightarrow\) \(\left(x+\sqrt{3}\right)^2-9=0\)
\(\Leftrightarrow\) \(\left(x+\sqrt{3}-3\right).\left(x+\sqrt{3}+3\right)=0\)
\(\Leftrightarrow\) \(\left[\begin{array}{} x+\sqrt{3}-3=0 \\ x+\sqrt{3}+3=0 \end{array} \right.\)\(\Leftrightarrow\) \(\left[\begin{array}{} x= 3-\sqrt{3} \\ x= -3-\sqrt{3} \end{array} \right.\)
Vậy phương trình có tập nghiệm là S={\(3-\sqrt{3};-3-\sqrt{3}\)}
\(\left\{{}\begin{matrix}\dfrac{8}{x-1}+\dfrac{15}{y+2}=1\\\dfrac{1}{x-1}+\dfrac{1}{y+2}=\dfrac{1}{12}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{8}{x-1}+\dfrac{15}{y+2}=1\\\dfrac{8}{x-1}+\dfrac{8}{y+2}=\dfrac{2}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{7}{y+2}=\dfrac{1}{3}\\\dfrac{1}{x-1}+\dfrac{1}{y+2}=\dfrac{1}{12}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y+2=21\\\dfrac{1}{x-1}=\dfrac{1}{12}-\dfrac{1}{21}=\dfrac{1}{28}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=19\\x=29\end{matrix}\right.\)
Bài 1:
\(P=\left(\dfrac{x-\sqrt{x}-2+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}}{\sqrt{x}-2}\right)\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)
\(=\dfrac{x-\sqrt{x}+2-x-\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)
\(=\dfrac{-2\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\cdot\dfrac{1}{\sqrt{x}-1}=\dfrac{-2}{\sqrt{x}+1}\)
\(\left\{{}\begin{matrix}x+y=m-1\\x-y=m+3\end{matrix}\right.\)
\(\Rightarrow x+y+x-y=m-1+m+3\)
\(\Rightarrow2x=2m+2\Rightarrow x=m+1\)
\(\Rightarrow x_0=m+1\) (1)
\(\left\{{}\begin{matrix}x+y=m-1\\x-y=m+3\end{matrix}\right.\)
\(\Rightarrow x+y-\left(x-y\right)=m-1-\left(m+3\right)\)
\(\Rightarrow2y=-4\Rightarrow y=-2\Rightarrow y_0=-2\Rightarrow y_0^2=4\) (2)
-Từ (1) và (2) suy ra:
\(m+1=4\Rightarrow m=3\)
Cái này mình biết chút... nhưng mà giải trên đây không tiện lắm bạn có chới zalo ko gửi ad qua cho mình để kp rồi mình gửi lời giải qua luôn...
\(\Leftrightarrow\left\{{}\begin{matrix}15\sqrt{x-2}+10\sqrt{y+1}=35\\15\sqrt{x-2}-9\sqrt{y+1}=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}19\sqrt{y+1}=38\\3\sqrt{x-2}+2\sqrt{y+1}=7\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{y+1}=2\\3\sqrt{x-2}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y+1=4\\x-2=1\end{matrix}\right.\Leftrightarrow\left(x,y\right)=\left(3;3\right)\)
\(\dfrac{3}{x+2}-1=\dfrac{5x+7}{x+2}\)
\(\Leftrightarrow\dfrac{3}{x+2}-\dfrac{x+2}{x+2}=\dfrac{5x+7}{x+2}\)
\(\Rightarrow3-x-2=5x+7\)
\(\Leftrightarrow-x-5x=7-3+2\)
\(\Leftrightarrow-6x=6\Rightarrow x=-1\)