Bài 1:

a) \(=\dfrac{\sqrt{5}.\sqrt{7}}{5}=\dfrac{\sqrt{35}}{5}\)

b) \(=\dfrac{\left|y\right|}{\sqrt{3}}=\dfrac{\sqrt{3}y}{3}\)

c) \(=\dfrac{\sqrt{2}}{\sqrt{t}}=\dfrac{\sqrt{2t}}{t}\)

d) \(=\sqrt{\dfrac{7p^2-3p^2}{7}}=\sqrt{\dfrac{4p^2}{7}}=\dfrac{2\left|p\right|}{\sqrt{7}}=\dfrac{-2\sqrt{7}p}{7}\)

Bài 2:

a) \(=\dfrac{\sqrt{21}-\sqrt{15}}{3}\)

b) \(=\dfrac{10\left(4+3\sqrt{2}\right)}{16-18}=-20-15\sqrt{2}\)

c) \(=\dfrac{\left(3\sqrt{10}-5\right)\left(6+\sqrt{10}\right)}{36-10}=\dfrac{18\sqrt{10}+30-30-5\sqrt{10}}{26}=\dfrac{13\sqrt{10}}{26}=\dfrac{\sqrt{10}}{2}\)

Còn câu d bài 2 thì sao ạ??

ta có sinB=\(\dfrac{AH}{AB}\)\(\Rightarrow\)AH=AB.sinB=3,6.sin62=3,18

BH=\(\sqrt{AB^2-AH^2}\)(pytago)=\(\sqrt{3,6^2-3,18^2}\)=1,69

\(_{\widehat{C}}\)=90-\(\widehat{B}\)=90-62=28\(^0\)

sinC=\(\dfrac{AB}{BC}\)\(\Rightarrow\)BC=\(\dfrac{AB}{sinC}\)=\(\dfrac{3,6}{sin28}\)=7,67

mà:CH=BC-BH=7,67-1,69=5,98

AC=\(\sqrt{BC^2-AB^2}\)(pytago)=\(\sqrt{7,67^2-3,6^2}\)=6.77

Lời giải:

a. ĐKXĐ: $x>0; x\neq 4$

b.

\(M=\sqrt{x}.\left[\frac{1}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}\right].\frac{x-4}{2\sqrt{x}}\)

\(=\frac{2\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}.\frac{x-4}{2}=\frac{2\sqrt{x}}{x-4}.\frac{x-4}{2}=\sqrt{x}\)

c. Để $M>3\Leftrightarrow \sqrt{x}>3\Leftrightarrow x>9$

Kết hợp đkxđ suy ra $x>9$ thì $M>3$

\(a,=2\sqrt{2}\left(\sqrt{5}-1\right)\sqrt{4+\sqrt{\left(\sqrt{5}-1\right)^2}}\\ =2\sqrt{2}\left(\sqrt{5}-1\right)\sqrt{4+\sqrt{5}-1}\\ =2\left(\sqrt{5}-1\right)\sqrt{6-2\sqrt{5}}\\ =2\left(\sqrt{5}-1\right)\sqrt{\left(\sqrt{5}-1\right)^2}\\ =2\left(\sqrt{5}-1\right)^2=2\left(6-2\sqrt{5}\right)=12-4\sqrt{5}\\ b,=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8-2\sqrt{15}}\\ =\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\\ =\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)^2\\ =\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\\ =32-8\sqrt{15}+8\sqrt{15}-30=2\)

Câu 4: ĐKXĐ: x>=1/2

Ta có: \(2\left(x-\sqrt{2x^2+5x-3}\right)=1+x\left(\sqrt{2x-1}-2\sqrt{x+3}\right)\)

=>\(2x-2\sqrt{2x^2+5x-3}=1+x\cdot\sqrt{2x-1}-2x\cdot\sqrt{x+3}\)

=>\(2x-1-2\cdot\sqrt{\left(2x-1\right)\left(x+3\right)}-x\cdot\sqrt{2x-1}+2x\cdot\sqrt{x+3}=0\)

=>\(2x-1-x\cdot\sqrt{2x-1}-2\cdot\sqrt{\left(2x-1\right)\left(x+3\right)}+2x\cdot\sqrt{x+3}=0\)

=>\(\sqrt{2x-1}\left(\sqrt{2x-1}-x\right)-2\cdot\sqrt{x+3}\left(\sqrt{2x-1}-x\right)=0\)

=>\(\left(\sqrt{2x-1}-\sqrt{4x+12}\right)\left(\sqrt{2x-1}-x\right)=0\)

TH1: \(\sqrt{2x-1}-\sqrt{4x+12}=0\)

=>\(\sqrt{2x-1}=\sqrt{4x+12}\)

=>4x+12=2x-1

=>2x=-13

=>\(x=-\frac{13}{2}\) (loại)

TH2: \(\sqrt{2x-1}-x=0\)

=>\(\sqrt{2x-1}=x\)

=>\(\begin{cases}2x-1=x^2\\ x\ge0\end{cases}\Rightarrow\begin{cases}x^2-2x+1=0\\ x\ge0\end{cases}\Rightarrow\begin{cases}\left(x-1\right)^2=0\\ x\ge0\end{cases}\)

=>x-1=0

=>x=1(nhận)

\(x^2-2\left(m-2\right)x+m^2+2m-3=0\left(1\right)\)

Để phương trình có hai nghiệm phân biệt thì Δ' > 0

\(\Rightarrow\left(m-2\right)^2-m^2-2m+3>0\Leftrightarrow m^2-4m+4-m^2-2m+3>0\Leftrightarrow-6m+7>0\Leftrightarrow m< \dfrac{7}{6}\)\)

Theo viét : \(\(\left\{{}\begin{matrix}x_1+x_2=2\left(m-2\right)\\x_1x_2=m^2+2m-3\end{matrix}\right.\)\)

Lại có :\( \dfrac{1}{x_1}+\dfrac{1}{x_2}=\dfrac{x_1+x_2}{5}\)

\(\Leftrightarrow\dfrac{x_1+x_2}{x_1x_2}=\dfrac{x_1+x_2}{5}\)

\(\Rightarrow\left(x_1+x_2\right)\left(x_1x_2\right)=5\left(x_1+x_2\right)\)

\(\Leftrightarrow\left(2m-4\right)\left(m^2+2m-3\right)=5\left(2m-4\right)\)

\(\Leftrightarrow2m^3+4m^2-6m-4m^2-8m+12=10m-20\)

\(\Leftrightarrow2m^3-24m+32=0\) \(\Leftrightarrow\left[{}\begin{matrix}m=-4\left(n\right)\\m=2\left(l\right)\end{matrix}\right.\)

Vậy \(m=-4\) thì thỏa điều kiện

bạn quy đồng chỗ  \(\dfrac{1}{x_1}+\dfrac{1}{x_2}=\dfrac{x_2}{x_1x_2}+\dfrac{x_1}{x_1x_2}=\dfrac{x_1+x_2}{x_1x_2}\)

\(BC=\sqrt{8^2+5^2}=\sqrt{89}\approx9,4\left(cm\right)\)

Bài 5: 

a: Xét ΔBEC và ΔADC có 

\(\widehat{C}\) chung

\(\widehat{EBC}=\widehat{DAC}\)

Do đó: ΔBEC\(\sim\)ΔADC

Giusp mình với mọi người ơi!!!