a) \(8\sqrt{x+2}\) + \(\sqrt{11-x}\) - \(2\sqrt{22+9x-x^2}\)+ 4 =0
b) \(\sqrt{1+4x}\)+ \(2\sqrt{2-x}\)+\(2\sqrt{\left(1+4x\right)\left(2-x\right)}\)=3
c) \(\sqrt{8+\sqrt{x}}\)+\(\sqrt{5-\sqrt{x}}\)=5
d) \(\sqrt{x^4-1}\)-2 =\(\sqrt{x-1}\)- \(2\sqrt{x^3+x^2+x+1}\)
\(\left(\sqrt{8-\sqrt{X-3}}-\sqrt{5-\sqrt{X-3}}\right)^2=5^2\)=\(5^2\)
\(\Leftrightarrow-2\sqrt{\left(8-\sqrt{X-3}\right)\left(5-\sqrt{X-3}\right)}=25-3=22\)
\(\Leftrightarrow-\sqrt{\left(8-\sqrt{X-3}\right)\left(5-\sqrt{X-3}\right)}=11\)
Do \(\sqrt{\left(8-\sqrt{X-3}\right)\left(5-\sqrt{X-3}\right)}\ge0\Rightarrow-\sqrt{\left(8-\sqrt{X-3}\right)\left(5-\sqrt{X-3}\right)}\le0\)
\(\Rightarrow\)PT vô nghiệm
Đk: \(x\ge\frac{2}{3}\)
Ta có: \(x^2+1^2\ge2x=\left(2x-1\right)+1=\left(\sqrt{2x-1}\right)^2+1^2\ge2\sqrt{2x-1}\left(1\right)\)
Lại có: \(\left(\sqrt{x}+\sqrt{3x-2}\right)^2\le2\left(x+3x-2\right)=2\left(4x-2\right)=4\left(2x-1\right)\)
suy ra: \(\sqrt{x}+\sqrt{3x-2}\le2\sqrt{2x-1}\left(2\right)\)
Từ (1);(2) suy ra \(x^2+1\ge\sqrt{x}+\sqrt{3x-2}\)
Để dấu"=" xảy ra theo đề bài thì x=1
\(\left(6\right)\dfrac{3\sqrt{x}}{5\sqrt{x}-1}\le-3\)
\(\left(7\right)\dfrac{8\sqrt{x}+8}{6\sqrt{x}+9}>\dfrac{8}{3}\)
\(\left(8\right)\dfrac{\sqrt{x}-2}{2\sqrt{x}-3}< -4\)
\(\left(9\right)\dfrac{4\sqrt{x}+6}{5\sqrt{x}+7}\le-\dfrac{2}{3}\)
\(\left(10\right)\dfrac{6\sqrt{x}-2}{7\sqrt{x}-1}>-6\)
Giải các phương trình sau:
a \(x^2-11=0\)
b \(x^2-12x+52=0\)
c \(x^2-3x-28=0\)
d \(x^2-11x+38=0\)
e \(6x^2+71x+175=0\)
f \(x^2-\left(\sqrt{2}+\sqrt{8}\right)x+4=0\)
g\(\left(1+\sqrt{3}\right)x^2-\left(2\sqrt{3}+1\right)x+\sqrt{3}=0\)
Bài 1: Rút gọn
a. \(\left(5-2\sqrt{3}\right)^2+\left(5+2\sqrt{3}\right)^2\)
b. \(\left(\sqrt{5}+\sqrt{2}\right)^2-\left(2\sqrt{5}+1\right)\left(2\sqrt{5}-1\right)-\sqrt{40}\)
c. \(\left(\sqrt{2}-1\right)^2-\frac{2}{3}\sqrt{4}+\frac{4\sqrt{2}}{5}+\sqrt{1\frac{11}{15}}-\sqrt{2}\)
d. \(\left(\sqrt{6}-\sqrt{18}+5\sqrt{2}-\frac{1}{2}\sqrt{8}\right)2\sqrt{6}+2\sqrt{3}\)
e. \(\left(2\sqrt{3}-3\sqrt{2}\right)^2+6\sqrt{6}+3\sqrt{24}\)
Bài 2: Rút gọn
A =\(\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}:\frac{\sqrt{x+1}}{x-2\sqrt{x}+1}\right)\)(x>0 ; x khác 1)
giải pt: \(11\sqrt{5-x}+8\sqrt{2x-1}=24+3\sqrt{\left(5-x\right)\left(2x-1\right)}\)
\(\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}
\)
\(\dfrac{x\sqrt{x}+1}{\sqrt{x}+1}\)
\(\left(1-\dfrac{4}{\sqrt{x}-1}+\dfrac{1}{x-1}\right):\left(\sqrt{x}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right)\)
\(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\)
TÌM ĐKXĐ VÀ RÚT GỌN
Giải phương trình:
a)\(\sqrt{\sqrt{5}-\sqrt{3x}}=\sqrt{8+2\sqrt{15}}\)
b)\(\sqrt{4x-20}-3\sqrt{\dfrac{x-5}{9}}=\sqrt{1-x}\)
c) \(\sqrt{4x+8}+2\sqrt{x+2}-\sqrt{9x+18}=1\)
d) \(\sqrt{x^2-6x+9}+x=11\)
e) \(\sqrt{3x^2-4x+3}=1-2x\)
f) \(\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}=4\)
g) \(\sqrt{9x+9}+\sqrt{4x+4}=\sqrt{x+1}\)
