\(\sqrt{x+2-4\sqrt{x-2}}+\sqrt{x+7-6\sqrt{x-2}}=1\)ĐK : x >= 2 

\(\Leftrightarrow\sqrt{x-2-4\sqrt{x-2}+4}+\sqrt{x-2-6\sqrt{x-2}+9}=1\)

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

\(\Leftrightarrow\sqrt{x-2}-2+\sqrt{x-2}-3=1\Leftrightarrow2\sqrt{x-2}=6\)

\(\Leftrightarrow\sqrt{x-2}=3\Leftrightarrow x-2=9\Leftrightarrow x=11\)(tmđk)

ĐK : x >= 2

\(\Leftrightarrow\sqrt{\left(x-2\right)-4\sqrt{x-2}+4}+\sqrt{\left(x-2\right)-6\sqrt{x-2}+9}=1\)

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

\(\Leftrightarrow\left|\sqrt{x-2}-2\right|+\left|\sqrt{x-2}-3\right|=1\)(1)

Với 2 ≤ x < 4 (1) trở thành \(2-\sqrt{x-2}+3-\sqrt{x-2}=1\Leftrightarrow-2\sqrt{x-2}=-4\Leftrightarrow\sqrt{x-2}=2\Leftrightarrow x=6\left(ktm\right)\)

Với 4 ≤ x < 11 (1) trở thành \(\sqrt{x-2}-2+3-\sqrt{x-2}=1\Leftrightarrow1=1\left(luondung\right)\)(2)

Với x ≥ 11 (1) trở thành \(\sqrt{x-2}-2+\sqrt{x-2}-3=1\Leftrightarrow2\sqrt{x-2}=6\Leftrightarrow\sqrt{x-2}=3\Leftrightarrow x=11\)( tm ) (3)

Từ (2) và (3) => S = { x | 4 ≤ x ≤ 11 }

a, ĐK : \(4-x^2\ge0\Leftrightarrow\left(2-x\right)\left(x+2\right)\ge0\Leftrightarrow-2\le x\le2\)

b, ĐK : \(16-x^2\ge0\Leftrightarrow\left(4-x\right)\left(x+4\right)\ge0\Leftrightarrow-4\le x\le4\)

c, ĐK : \(x^2-3\ge0\Leftrightarrow\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)\ge0\Leftrightarrow x\le-\sqrt{3};x\ge\sqrt{3}\)

\(A=\left|3-0,7.9\right|+\frac{2}{5}=\left|3-2,7\right|+0,4=\left|0,3\right|+0,4=0,3+0,4=0,7\)

\(B=\left|x\right|+x-5=-x+x-5=-5\)

\(C=\left|2-x\right|+\left|x-3\right|=x-2+3-x=1\)

\(\hept{\begin{cases}3x^2-5xy-4y^2=-3\\-8x^2+11xy+9y^2=6\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}6x^2-10xy-8y^2=-6\\-8x^2+11xy+9y^2=6\end{cases}}\)

\(\Rightarrow-2x^2+xy+y^2=0\)

\(\Leftrightarrow-2x^2+2xy-xy+y^2=0\)

\(\Leftrightarrow-2x\left(x-y\right)-y\left(x-y\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left(-2x-y\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=y\\y=2x\end{cases}}\)

th1 : x = y  ta có :

3x^2 - 5x^2 - 4x^2 = -3

=>-6x^2 = -3

=> x^2 = 1/2

=> x = +- căn 1/2 = y

th2 : y =2x ta có 

3x^2 - 10x^2 - 4(2x)^2 = -3

=> 3x^2 - 10x^2 - 16x^2 = -3

=> -23x^2 = -3

=> x^2 = 3/23

=> x = +- căn 3/23 

tính nốt y đi

viết này thì sao biết được bạn

a, \(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)

\(=\sqrt{4+\sqrt{15}}\sqrt{4-\sqrt{15}}\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4+\sqrt{15}}\)

\(=\sqrt{16-15}\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)\sqrt{4+\sqrt{15}}\)

\(=\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8+2\sqrt{5.3}}=\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}\)

\(=\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)=5-3=2\)

làm hộ 1 câu thôi

mấy câu kia cũng kiểu kiểu thế á b

-))

Với \(x>0;x\ne4;9\)

\(C=\left(\frac{\sqrt{x}+1}{\sqrt{x}-2}-\frac{2\sqrt{x}}{\sqrt{x}+2}+\frac{5\sqrt{x}+2}{4-x}\right):\frac{3\sqrt{x}-x}{x+4\sqrt{x}+4}\)

\(=\left(\frac{x+3\sqrt{x}+2-2x+4\sqrt{x}-2-5\sqrt{x}}{x-4}\right):\frac{\sqrt{x}\left(3-\sqrt{x}\right)}{\left(\sqrt{x}+2\right)^2}\)

\(=\left(\frac{-x+2\sqrt{x}}{x-4}\right):\frac{\sqrt{x}\left(3-\sqrt{x}\right)}{\left(\sqrt{x}+2\right)^2}\)

\(=\frac{-\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)^2}{\left(x-4\right)\sqrt{x}\left(3-\sqrt{x}\right)}=\frac{\sqrt{x}+2}{\sqrt{x}-3}\)

\(A=\left|2x+1\right|+5\)

Ta có: \(\left|2x+1\right|\ge0,\forall x\)

\(\Rightarrow\left|2x+1\right|+5\ge5,\forall x\)

Dấu "\(=\)" xảy ra \(\Leftrightarrow\left|2x+1\right|=0\)

\(\Leftrightarrow2x+1=0\)

\(\Leftrightarrow2x=-1\)

\(\Leftrightarrow x=\frac{-1}{2}\)

Vậy, Giá trih nhỏ nhất\(minA=5\)\(\Leftrightarrow x=\frac{-1}{2}\)

t.ick và chọn câu trả lời của mình nhé

Chúc bạn học tốt!

A = |2x - 1| + 5

có |2x - 1| ≥ 0 => |2x - 1| + 5 ≥ 5

=> A ≥ 5

xét A = 5 <=> |2x - 1| = 0 <=> x = 1/2

vậy_

B = 3 - |1 - x|

có |1-x| ≥ 0 => -|1 - x|  ≤ 0

=> 3 - |1 - x| ≤ 3

=> B ≤ 3

xét  B = 3 <=> |1-x| = 0 <=> x = 1

vậy_