\(ĐK:x\ge2;y\le1;z\ge-3\)

\(4x-y+z+10=4\sqrt{x-2}+6\sqrt{1-y}+4\sqrt{z+3}\)

\(\Leftrightarrow4x-y+z+10-4\sqrt{x-2}-6\sqrt{1-y}-4\sqrt{z+3}=0\)

\(\Leftrightarrow\left(4x-8-4\sqrt{x-2}+1\right)+\left(9-6\sqrt{1-y}+1-y\right)+\left(z+3-4\sqrt{z+3}+4\right)=0\)

\(\Leftrightarrow\left(2\sqrt{x-2}-1\right)^2+\left(3-\sqrt{1-y}\right)^2+\left(\sqrt{z+3}-2\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}2\sqrt{x-2}-1=0\\3-\sqrt{1-y}=0\\\sqrt{z+3}-2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{9}{4}\\y=-8\\z=1\end{cases}}\left(tm\right)\)

\(\Rightarrow4x+y+z=4\cdot\frac{9}{4}-8+1=2\)

a > 0 thì nó là ĐK luôn rồi bạn

\(A=\frac{\sqrt{a}\left(a\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\frac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1\)

\(=\frac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-2\sqrt{a}-1+1\)

\(=a+\sqrt{a}-2\sqrt{a}=a-\sqrt{a}\)

a, ĐK : \(x\ge0;x\ne9\)

b, \(Q=\frac{2\sqrt{x}}{\sqrt{x}+3}-\frac{\sqrt{x}-1}{3-\sqrt{x}}-\frac{3-11\sqrt{x}}{x-9}\)

\(=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)-3+11\sqrt{x}}{x-9}\)

\(=\frac{2x-6\sqrt{x}+x+2\sqrt{x}-3-3+11\sqrt{x}}{x-9}=\frac{3x+7\sqrt{x}-6}{x-9}=\frac{3\sqrt{x}-2}{\sqrt{x}-3}\)

a, ĐK :  \(x\ge0;x\ne4\)

b, \(P=\frac{2x-3\sqrt{x}-2}{\sqrt{x}-2}=\frac{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}{\sqrt{x}-2}=2\sqrt{x}+1\)

Tham khảo câu a , c

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Gọi 2 điểm cố định là \(A\left(x_0;y_0\right)\)

Thay vào ptđt (d) ta được : \(y_0=mx_0+m+1\Leftrightarrow mx_0+m+1-y_0=0\)

\(\Leftrightarrow m\left(x_0+1\right)+\left(1-y_0\right)=0\)

Đẳng thức xảy ra khi \(\hept{\begin{cases}x_0+1=0\\1-y_0=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x_0=-1\\y_0=1\end{cases}}\Rightarrow A\left(-1;1\right)\)

Vậy d luôn đi qua 1 điểm cố định A(-1;1) 

Hoành độ giao điểm của (P) và (d) là nghiệm của phương trình :

14.x2=x−114.x2=x−1

<=> x² = 4x - 4

<=> x² - 4x + 4 = 0 <=> (x - 2)² = 0 <=> x - 2= 0 <=> x = 2

=> y = 2-1 = 1

Vậy (P) cắt (d) tại 1 điểm duy nhất là (2;1) 

=> đpcm 

đúng ko ????????????? 

sai thì cho mik xin lỗi

Ta có \(2x^2\ge0\forall x\)

=> \(2x^2+9\ge9>0\)

=> Phương trình vô nghiệm 

2x² = 0 - 9 = -9

x² = -9 : 2 = 4,5

x= \(\sqrt{4,5}\)

a, Ta có: \(x=4-2\sqrt{3}\)\(=3-2\sqrt{3}+1\)\(=\left(\sqrt{3}-1\right)^2\)

         \(\Rightarrow\sqrt{x}=\sqrt{\left(\sqrt{3}-1\right)^2}\)\(=\sqrt{3}-1\)

Thay \(\sqrt{x}=\sqrt{3}-1\) vào biểu thức P ta có:

\(P=\frac{\sqrt{3}-1+1}{\sqrt{3}-1-4}\)\(=\frac{\sqrt{3}}{\sqrt{3}-5}\)\(=\frac{\sqrt{3}.\left(\sqrt{3}+5\right)}{\left(\sqrt{3}-5\right).\left(\sqrt{3}+5\right)}\)\(=\frac{3-5\sqrt{3}}{3-25}\)\(=\frac{5\sqrt{3}-3}{22}\)

Vậy \(P=\frac{5\sqrt{3}-3}{22}\)khi \(x=4-2\sqrt{3}\) 

b, \(E=\frac{1}{\sqrt{3}-1}-\frac{1}{\sqrt{3}+1}\)\(=\frac{\sqrt{3}+1}{\left(\sqrt{3}-1\right).\left(\sqrt{3}+1\right)}\)\(-\frac{\sqrt{3}-1}{\left(\sqrt{3}+1\right).\left(\sqrt{3}-1\right)}\)

       \(=\frac{\sqrt{3}+1-\sqrt{3}+1}{3-1}\)     \(=\frac{2}{2}=1\)

a, Ta có : \(x=4-2\sqrt{3}\Rightarrow\sqrt{x}=\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)

Thay vào P ta được : \(P=\frac{\sqrt{3}-1+1}{\sqrt{3}-1-4}=\frac{\sqrt{3}}{\sqrt{3}-5}=\frac{\sqrt{3}\left(\sqrt{3}+5\right)}{-22}=-\frac{3+5\sqrt{3}}{22}\)

b, \(E=\frac{1}{\sqrt{3}-1}-\frac{1}{\sqrt{3}+1}=\frac{\sqrt{3}+1-\sqrt{3}+1}{2}=1\)