câu a là tính tỉ số x/y nhé

\(x+y=3\sqrt{xy}\)

\(\Leftrightarrow\)\(\frac{x}{y}+1=3\sqrt{\frac{x}{y}}\)

\(\Leftrightarrow\)\(\frac{x}{y}-3\sqrt{\frac{x}{y}}+\frac{9}{4}=\frac{5}{4}\)

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

\(\Leftrightarrow\)\(\frac{x}{y}=\frac{7+3\sqrt{5}}{2}\)

Câu 1:

\(2\sqrt{\dfrac{3}{20}}+\sqrt{\dfrac{1}{60}}-\sqrt{\dfrac{1}{15}}\)

= \(\sqrt{\dfrac{2^2\cdot3}{20}}+\sqrt{\dfrac{1}{60}}-\sqrt{\dfrac{1}{15}}\)

= \(\sqrt{\dfrac{12}{20}}+\sqrt{\dfrac{1}{60}}-\sqrt{\dfrac{1}{15}}\)

= \(\dfrac{\sqrt{12}\cdot\sqrt{20}}{\left(\sqrt{20}\right)^2}+\dfrac{\sqrt{60}}{\left(\sqrt{60}\right)^2}-\dfrac{\sqrt{15}}{\left(\sqrt{15}\right)^2}\)

= \(\dfrac{\sqrt{240}}{20}+\dfrac{\sqrt{60}}{60}-\dfrac{\sqrt{15}}{15}\)

= \(\dfrac{\sqrt{15}}{5}+\dfrac{\sqrt{15}}{30}-\dfrac{\sqrt{15}}{15}\)

= \(\sqrt{15}\cdot\left(\dfrac{1}{5}+\dfrac{1}{30}-\dfrac{1}{15}\right)\)

= \(\sqrt{15}\cdot\dfrac{1}{6}\) = \(\dfrac{\sqrt{15}}{6}\)

Bài 2:

a)\(\dfrac{1}{\sqrt{18}+\sqrt{8}-2\sqrt{2}}=\dfrac{1}{\sqrt{18}+2\sqrt{2}-2\sqrt{2}}=\dfrac{1}{\sqrt{18}}=\dfrac{\sqrt{18}}{18}=\dfrac{\sqrt{2}}{6}\)

b)\(\dfrac{\sqrt{2}}{1+\sqrt{2}-\sqrt{3}}=\dfrac{\sqrt{2}\cdot\left(1+\sqrt{2}+\sqrt{3}\right)}{\left(1+\sqrt{2}\right)^2-3}=\dfrac{\sqrt{2}\cdot\left(1+\sqrt{2}+\sqrt{3}\right)}{1+2\sqrt{2}+2-3}=\dfrac{\sqrt{2}\cdot\left(1+\sqrt{2}+\sqrt{3}\right)}{2\sqrt{2}}=\dfrac{1}{2}\cdot\left(1+\sqrt{2}+\sqrt{3}\right)\)c) \(\dfrac{1}{\sqrt{3}+\sqrt{2}-\sqrt{5}}=\dfrac{\sqrt{3}+\sqrt{2}+\sqrt{5}}{\left(\sqrt{3}+\sqrt{2}\right)^2-5}=\dfrac{\sqrt{3}+\sqrt{2}+\sqrt{5}}{3+2\sqrt{6}+2-5}=\dfrac{\sqrt{3}+\sqrt{2}+\sqrt{5}}{2\sqrt{6}}=\dfrac{\sqrt{6}\cdot\left(\sqrt{3}+\sqrt{2}+\sqrt{5}\right)}{2\left(\sqrt{6}\right)^2}=\dfrac{\sqrt{6}}{12}\cdot\left(\sqrt{2}+\sqrt{3}+\sqrt{5}\right)\)

a,    ta có  

        \(\sqrt{8}+\sqrt{15}< \sqrt{9}+\sqrt{16}< 3+4< 7\)             (1)

lại có         \(\sqrt{65}-1>\sqrt{64}-1>8-1>7\)                 (2)

từ (1) và(2) =>\(\sqrt{8}+\sqrt{15}< \sqrt{65}-1\)

bài 2 

\(M=\sqrt{\frac{\left(2^3\right)^{10}-\left(2^2\right)^{10}}{\left(2^2\right)^{11}-\left(2^3\right)^4}}=\sqrt{\frac{2^{30}-2^{20}}{2^{22}-2^{12}}}=\sqrt{\frac{2^{20}\left(2^{10}-1\right)}{2^{12}\left(2^{10}-1\right)}}=\sqrt{\frac{2^{20}}{2^{12}}}=\sqrt{2^8}=2^4\)

Ta có: VT =82−32−41−2=1−2​82​−32​−4​

=82−2.42−41−2=82−42−41−2=1−2​82​−2.42​−4​=1−2​82​−42​−4​

=42−41−2=−4(1−2)1−2=−4==1−2​42​−4​=1−2​−4(1−2​)​=−4= V P

Vậy 82−32−41−2=−41−2​82​−32​−4​=−4

b) ĐKXĐ: {�≥0�+2≠0�−2≠0�−4≠0⇔{�≥0�≠2�≠4⇔{�≥0�≠4⎩⎨⎧​x≥0x​+2=0x​−2=0x−4=0​⇔⎩⎨⎧​x≥0x​=2x=4​⇔{x≥0x=4​.

Vậy ĐKXĐ của $P$ là $x\ge 0$, $x\ne 4$.

Với $x\ge 0$, $x\ne 4$ ta có:

�=(2�+2−1�−2+7�−4).(�−1)P=(x​+22​−x​−21​+x−47​).(x​−1)

=(2�+2−1�−2+7(�−2)(�+2)).(�−1)=(x​+22​−x​−21​+(x​−2)(x​+2)7​).(x​−1)

=(2(�−2)−(�+2)+7(�−2)(�+2)).(�−1)=((x​−2)(x​+2)2(x​−2)−(x​+2)+7​).(x​−1)

=2�−4−�−2+7(�−2)(�+2).(�−1)=(x​−2)(x​+2)2x​−4−x​−2+7​.(x​−1)

=�+1(�−2)(�+2).(�−1)=(x​−2)(x​+2)x​+1​.(x​−1)

=�−1�−4=x−4x−1​.

Vậy �=�−1�−4P=x−4x−1​ với $x\ge 0$, $x\ne 4$.

a) \(\dfrac{8\sqrt{2}-\sqrt{32}-4}{1-\sqrt{2}}=\dfrac{8\sqrt{2}-4\sqrt{2}-4}{1-\sqrt{2}}\)

\(=\dfrac{4\sqrt{2}-4}{1-\sqrt{2}}=\dfrac{-4\left(1-\sqrt{2}\right)}{1-\sqrt{2}}=-4\)

b) ĐKXĐ: \(x>0;x\ne4\)

\(P=\left(\dfrac{2}{\sqrt{x}+2}-\dfrac{1}{\sqrt{x}-2}+\dfrac{7}{x-4}\right).\left(\sqrt{x}-1\right)\)

\(P=\dfrac{2\sqrt{x}-4-\sqrt{x}-2+7}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}.\left(\sqrt{x}-1\right)\)

\(P=\dfrac{\sqrt{x}+1}{x-4}\left(\sqrt{x}-1\right)\)

\(P=\dfrac{x-1}{x-4}\)

a, 8

Bài 1:
Vì $x+y+z=1$ nên:

\(Q=\frac{x}{x+\sqrt{x(x+y+z)+yz}}+\frac{y}{y+\sqrt{y(x+y+z)+xz}}+\frac{z}{z+\sqrt{z(x+y+z)+xy}}\)

\(Q=\frac{x}{x+\sqrt{(x+y)(x+z)}}+\frac{y}{y+\sqrt{(y+z)(y+x)}}+\frac{z}{z+\sqrt{(z+x)(z+y)}}\)

Áp dụng BĐT Bunhiacopxky:

\(\sqrt{(x+y)(x+z)}=\sqrt{(x+y)(z+x)}\geq \sqrt{(\sqrt{xz}+\sqrt{xy})^2}=\sqrt{xz}+\sqrt{xy}\)

\(\Rightarrow \frac{x}{x+\sqrt{(x+y)(x+z)}}\leq \frac{x}{x+\sqrt{xy}+\sqrt{xz}}=\frac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)

Hoàn toàn tương tự với các phân thức còn lại và cộng theo vế suy ra:

\(Q\leq \frac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}+ \frac{\sqrt{y}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}+ \frac{\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}=1\)

Vậy $Q$ max bằng $1$

Dấu bằng xảy ra khi $x=y=z=\frac{1}{3}$

Bài 2:
Vì $x+y+z=1$ nên:

\(\text{VT}=\frac{1-x^2}{x(x+y+z)+yz}+\frac{1-y^2}{y(x+y+z)+xz}+\frac{1-z^2}{z(x+y+z)+xy}\)

\(\text{VT}=\frac{(x+y+z)^2-x^2}{(x+y)(x+z)}+\frac{(x+y+z)^2-y^2}{(y+z)(y+x)}+\frac{(x+y+z)^2-z^2}{(z+x)(z+y)}\)

\(\text{VT}=\frac{(y+z)[(x+y)+(x+z)]}{(x+y)(x+z)}+\frac{(x+z)[(y+z)+(y+x)]}{(y+z)(y+x)}+\frac{(x+y)[(z+x)+(z+y)]}{(z+x)(z+y)}\)

Áp dụng BĐT AM-GM:
\(\text{VT}\geq \frac{2(y+z)\sqrt{(x+y)(x+z)}}{(x+y)(x+z)}+\frac{2(x+z)\sqrt{(y+z)(y+x)}}{(y+z)(y+x)}+\frac{2(x+y)\sqrt{(z+x)(z+y)}}{(z+x)(z+y)}\)

\(\Leftrightarrow \text{VT}\geq 2\underbrace{\left(\frac{y+z}{\sqrt{(x+y)(x+z)}}+\frac{x+z}{\sqrt{(y+z)(y+x)}}+\frac{x+y}{\sqrt{(z+x)(z+y)}}\right)}_{M}\)

Tiếp tục AM-GM cho 3 số trong ngoặc lớn, suy ra \(M\geq 3\)

Do đó: \(\text{VT}\geq 2.3=6\) (đpcm)

Dấu bằng xảy ra khi $3x=3y=3z=1$

Bài 4:

Ta có một đẳng thức quen thuộc là:

\(1=(a+b)(b+c)(c+a)=(ab+bc+ac)(a+b+c)-abc(*)\)

Mà theo AM-GM:

\((a+b+c)(ab+bc+ac)\geq 3\sqrt[3]{abc}.3\sqrt[3]{ab.bc.ac}=9abc\)

\(\Rightarrow abc\leq \frac{(a+b+c)(ab+bc+ac)}{9}(**)\)

Từ \((*);(**)\Rightarrow 1\geq \frac{8}{9}(a+b+c)(ab+bc+ac)\)

Theo tính chất quen thuộc của BĐT AM-GM:

\((a+b+c)^2\geq 3(ab+bc+ac)\Rightarrow a+b+c\geq \sqrt{3(ab+bc+ac)}\)

Do đó:

\(1\geq \frac{8}{9}\sqrt{3(ab+bc+ac)^3}\)

\(\Rightarrow (ab+bc+ac)^3\leq \frac{27}{64}\Rightarrow ab+bc+ac\leq \frac{3}{4}\)

Ta có đpcm

Câu 1 :

\(P=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)

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

\(=\dfrac{2\sqrt{x}-9-x+9+2x-3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

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

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)

Câu 2 :

Ta có :

\(\Delta=m^2+16>0\)

\(=>\) phương trình có 2 nghiệm phân biệt .

Theo định lý vi-ét ta có :

\(\left\{{}\begin{matrix}x_1+x_2=m\\x_1.x_2=-4\end{matrix}\right.\)

Thay vào ta được :

\(\dfrac{2m+7}{m^2+8}\ge-\dfrac{1}{8}\)

\(\Leftrightarrow16m+56\ge-m^2-8\)

\(\Leftrightarrow m^2+16m+64\ge0\)

\(\Leftrightarrow\left(m+8\right)^2\ge0\) ( đúng )

1
a,A=\(\left(\sqrt{45}-\sqrt{20}+\sqrt{5}\right):\sqrt{6}\)
A=\(\left(3\sqrt{5}-2\sqrt{5}+\sqrt{5}\right):\sqrt{6}\)
A=\(2\sqrt{5}:\sqrt{6}=\dfrac{2\sqrt{5}}{\sqrt{6}}=\dfrac{\sqrt{30}}{3}\)
b, B=\(\dfrac{\sqrt{10}-\sqrt{15}}{\sqrt{8}-\sqrt{12}}=\dfrac{\sqrt{5.2}-\sqrt{5.3}}{\sqrt{4.2}-\sqrt{4.3}}=\dfrac{\sqrt{5}\left(\sqrt{2}-\sqrt{3}\right)}{\sqrt{4}\left(\sqrt{2}-\sqrt{3}\right)}\)
B=\(\dfrac{\sqrt{5}}{2}\)

Câu 1)

a) \(\left(\sqrt{45}-\sqrt{20}+\sqrt{5}\right):\sqrt{6}=\left(\sqrt{9.5}-\sqrt{4.5}+\sqrt{5}\right):\sqrt{6}=\left(3\sqrt{5}-2\sqrt{5}+\sqrt{5}\right):\sqrt{6}=\dfrac{2\sqrt{5}}{\sqrt{6}}=\dfrac{\sqrt{30}}{3}\)

b) \(\dfrac{\sqrt{10}-\sqrt{15}}{\sqrt{8}-\sqrt{12}}=\dfrac{\sqrt{5}\left(\sqrt{2}-\sqrt{3}\right)}{\sqrt{4}\left(\sqrt{2}-\sqrt{3}\right)}=\dfrac{\sqrt{5}}{\sqrt{4}}=\dfrac{\sqrt{5}}{2}\)

Câu 2)

ĐK: x\(\ge5\)

\(\sqrt{x-5}+\sqrt{4x-20}-\dfrac{1}{5}\sqrt{9x-45}=3\Leftrightarrow\sqrt{x-5}+\sqrt{4\left(x-5\right)}-\dfrac{1}{5}\sqrt{9\left(x-5\right)}=3\Leftrightarrow\sqrt{x-5}+2\sqrt{x-5}-\dfrac{3}{5}\sqrt{x-5}=3\Leftrightarrow\dfrac{12}{5}\sqrt{x-5}=3\Leftrightarrow\sqrt{x-5}=\dfrac{5}{4}\Leftrightarrow x-5=\dfrac{25}{16}\Leftrightarrow x=\dfrac{105}{16}\left(tm\right)\)

2, ĐK: x\(\ge5\) Ta được :
\(\sqrt{x-5}+\sqrt{4\left(x-5\right)}-\dfrac{1}{5}\sqrt{9\left(x-5\right)}=3\)
\(\Leftrightarrow\sqrt{x-5}+2\sqrt{x-5}-\dfrac{3}{5}\sqrt{x-5}=3\)
\(\Leftrightarrow\dfrac{4}{3}\sqrt{x-5}=3\)
\(\Leftrightarrow\sqrt{x-5}=\dfrac{9}{4}\)
\(\Leftrightarrow x-5=\dfrac{81}{16}\)
\(\Leftrightarrow x=\dfrac{161}{16}\left(tm\right)\)

giúp mình câu 1 trước đi nè

Câu 2:

a, ĐKXĐ: x\(\ge\)0; x\(\ne\)\(\pm\)1

B=

\(\left(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}}{\sqrt{x}+1}\\ =\dfrac{\left(\sqrt{x}-1\right)^2-\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}}{\sqrt{x}+1}\\ =\dfrac{-2.2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}}{\sqrt{x}+1}\\ =\dfrac{-4\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}}{\sqrt{x}+1}\\ =\dfrac{4\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\dfrac{\sqrt{x}+1}{\sqrt{x}}\\ =-\dfrac{4}{\sqrt{x}-1}\)

Câu 1: 

a: \(=\dfrac{-5+2\sqrt{6}+5+2\sqrt{6}}{1}=4\sqrt{6}\)

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

\(=-\sqrt{3}-\sqrt{5}+\sqrt{3}=-\sqrt{5}\)

c: \(=\dfrac{\sqrt{2}+1}{\sqrt{3}+\sqrt{2}}+\dfrac{2}{\sqrt{5}+\sqrt{3}}\)

\(=\left(\sqrt{2}+1\right)\left(\sqrt{3}-\sqrt{2}\right)+\sqrt{5}-\sqrt{3}\)

\(=\sqrt{6}-2+\sqrt{3}-\sqrt{2}+\sqrt{5}-\sqrt{3}\)

\(=\sqrt{6}-2-\sqrt{2}+\sqrt{5}\)

Câu 1 =3/10

\(1,\sqrt{\left(-0,3\right)^2}=\sqrt{0,09}=0,3\)

\(2,-\frac{1}{2}\sqrt{\left(0,3\right)^2}=-\frac{1}{2}.0,3=-0,15\)

\(3,\sqrt{a^{10}}=\sqrt{\left(a^5\right)^2}=a^5\left(a\ge0\right)\)

\(4,\sqrt{\left(2-x\right)^2}=\left|2-x\right|=2-x\left(x\le2\right)\)

\(5,\sqrt{x^2+2x+1}=\sqrt{\left(x+1\right)^2}=\left|x+1\right|\)

\(6,\sqrt{\left(1-\sqrt{2}\right)^2}=\left|1-\sqrt{2}\right|=\sqrt{2}-1\)(Vì \(1< \sqrt{2}\))

\(7,\sqrt{11+6\sqrt{2}}=\sqrt{9+6\sqrt{2}+2}=\sqrt{\left(3+\sqrt{2}\right)^2}=3+\sqrt{2}\)

\(8,\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}=\sqrt{7-2\sqrt{7}+1}-\sqrt{7+2\sqrt{7}+1}\)

                                                                    \(=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}\)

                                                                    \(=\left(\sqrt{7}-1\right)-\left(\sqrt{7}+1\right)\)

                                                                      \(=-2\)

\(9,\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}=\sqrt{5+2\sqrt{5}+1}+\sqrt{5-2\sqrt{5}+1}\)

                                                                    \(=\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}\)

                                                                    \(=\sqrt{5}+1+\sqrt{5}-1\)

                                                                    \(=2\sqrt{5}\)

\(\sqrt{\left(-0,3\right)^2}\)

\(=\left|0,3\right|\)

\(=0,3\)