Bài 1: 

a: ĐKXĐ: \(x\ge\dfrac{3}{2}\)

Bài 4:

\(a,A=\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\\ P=A:B=\dfrac{\sqrt{x}+1}{\sqrt{x}}\cdot\dfrac{x-1}{\sqrt{x}+1}=\dfrac{x-1}{\sqrt{x}}\\ b,P\sqrt{x}=m-\sqrt{x}+x\\ \Leftrightarrow x-1=m-\sqrt{x}+x\\ \Leftrightarrow m=\sqrt{x}-1\)

Bài 3:

b: Xét ΔABC có

I là trung điểm của BC

IK//AC

Do đó: K là trung điểm của AB

Xét ΔABC có

I là trung điểm của BC

IH//AB

Do đó: H là trung điểm của AC

Xét ΔABC có 

K là trung điểm của AB

H là trung điểm của AC

Do đó: HK là đường trung bình của ΔABC

Suy ra: HK//BC

jimmmmmmmmmmmmmmmmmmmmmmmmmmm

he he he he he he

bài 1:

bn lấy giá trị của √(4^2-3,9^2) là dc

bài 2

AB+BC=2√(3^2+4^2)=??

\(R_{tđ}=\dfrac{R_1\cdot R_2}{R_1+R_2}=\dfrac{24\cdot12}{24+12}=8\Omega\)

\(I=\dfrac{U}{R}=\dfrac{12}{8}=1,5A\)

\(P=\dfrac{U^2}{R}=\dfrac{12^2}{8}=18W\)

\(Q_{tỏa1}=A_1=U_1\cdot I_1\cdot t=12\cdot\dfrac{12}{24}\cdot1\cdot3600=21600J\)

\(Q_{tỏa2}=A_2=U_2\cdot I_2\cdot t=12\cdot\dfrac{12}{12}\cdot1\cdot3600=43200J\)

\(\dfrac{9^{15}.8^{11}}{3^{29}.16^8}=\dfrac{\left(3^2\right)^{15}.\left(2^3\right)^{11}}{3^{29}.\left(2^4\right)^8}=\dfrac{3^{30}.2^{33}}{3^{29}.2^{32}}\)

Ta lấy vễ trên chia vế dưới

\(=3.2=6\)

\(\dfrac{2^{11}.9^3}{3^5.16^2}=\dfrac{2^{11}.\left(3^2\right)^3}{3^5.\left(2^4\right)^2}=\dfrac{2^{11}.3^6}{3^5.2^8}\)

Ta lấy vế trên chia vế dưới

\(=2^3.3=24\)

\(\dfrac{9^{15}.8^{11}}{3^{29}.16^8}=\dfrac{\left(3^2\right)^{15}.\left(2^3\right)^{11}}{3^{29}.\left(2^4\right)^8}=\dfrac{3^{30}.2^{33}}{3^{29}.3^{32}}=3.2=6\)
\(\dfrac{2^{11}.9^3}{3^5.16^2}=\dfrac{2^{11}.\left(3^2\right)^3}{3^5.\left(2^4\right)^2}=\dfrac{2^{11}.3^6}{3^5.2^8}=2^3.3=8.3=24\)

\(\dfrac{9^{15}.8^{11}}{3^{29}.16^8}=\dfrac{3^{30}.2^{33}}{3^{29}.2^{32}}=3.2=6\)

\(\dfrac{2^{11}.9^3}{3^5.16^2}=\dfrac{2^{11}.3^6}{3^5.2^8}=2^3.3=8.3=24\)

cot B = \(\dfrac{5}{13}=>tanB=\dfrac{13}{5}\)
AC=AB.tanB
AC= 15.\(\dfrac{13}{5}\)
AC= 39cm
BC²=AB²+AC²
BC²=225+1521=1746
BC=3 \(\sqrt{194}\)

Bài 3: 

a) Ta có: \(\dfrac{a+4\sqrt{a}+4}{\sqrt{a}+2}-\dfrac{a-4}{\sqrt{a}-2}\)

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

=0

b) Ta có: \(\dfrac{9-a}{\sqrt{a}+3}-\dfrac{a-6\sqrt{a}+9}{\sqrt{a}-3}\)

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

\(=6-2\sqrt{a}\)

c) Ta có: \(\dfrac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\dfrac{a-b}{\sqrt{a}+\sqrt{b}}\)

\(=\sqrt{a}-\sqrt{b}-\left(\sqrt{a}-\sqrt{b}\right)\)

=0

d) Ta có: \(\dfrac{a\sqrt{a}-b\sqrt{b}}{\sqrt{a}-\sqrt{b}}+\sqrt{ab}\)

\(=a+\sqrt{ab}+b+\sqrt{ab}\)

\(=a+2\sqrt{ab}+b\)

Bài 1:

a.

\(\frac{4\sqrt{5}+\sqrt{15}}{\sqrt{5}}=\frac{\sqrt{5}(4+\sqrt{3})}{\sqrt{5}}=4+\sqrt{3}\)

$\frac{7-\sqrt{7}}{3\sqrt{7}}=\frac{\sqrt{7}(\sqrt{7}-1)}{3\sqrt{7}}=\frac{\sqrt{7}-1}{3}$
\(\frac{4\sqrt{2}-\sqrt{6}}{2\sqrt{3}}=\frac{\sqrt{2}(4-\sqrt{3})}{\sqrt{2}.\sqrt{6}}=\frac{4-\sqrt{3}}{\sqrt{6}}\)

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

b.

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

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

\(\frac{a+10\sqrt{a}+25}{\sqrt{a}+5}=\frac{(\sqrt{a}+5)^2}{\sqrt{a}+5}=\sqrt{a}+5\)

\(\frac{a-9}{\sqrt{a}+3}=\frac{(\sqrt{a}-3)(\sqrt{a}+3)}{\sqrt{a}-3}=\sqrt{a}+3\)

Bài 2.

a.

\(\frac{6-\sqrt{6}}{\sqrt{6}-1}+\frac{6+\sqrt{6}}{\sqrt{6}}=\frac{\sqrt{6}(\sqrt{6}-1)}{\sqrt{6}-1}+\frac{\sqrt{6}(\sqrt{6}+1)}{\sqrt{6}}=\sqrt{6}+(\sqrt{6}+1)=2\sqrt{6}+1\)

b.

\(\frac{3+\sqrt{3}}{\sqrt{3}}+\frac{\sqrt{6}-\sqrt{3}}{1-\sqrt{2}}=\frac{\sqrt{3}(\sqrt{3}+1)}{\sqrt{3}}+\frac{-\sqrt{3}(1-\sqrt{2})}{1-\sqrt{2}}=\sqrt{3}+1-\sqrt{3}=1\)

c.

\(\frac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}+\frac{6+2\sqrt{6}}{\sqrt{3}+\sqrt{2}}=\frac{\sqrt{3}(\sqrt{5}-2)}{\sqrt{5}-2}+\frac{\sqrt{6}(\sqrt{6}+2)}{\sqrt{3}+\sqrt{2}}\)

\(=\sqrt{3}+\frac{\sqrt{6}.\sqrt{2}(\sqrt{3}+\sqrt{2})}{\sqrt{3}+\sqrt{2}}=\sqrt{3}+\sqrt{12}=3\sqrt{3}\)

\(h=2R\)

\(V=h.\pi R^2=2R.\pi R^2=16\pi\)

\(\Rightarrow R^3=8\Rightarrow R=2\Rightarrow h=4\)

\(S_{tp}=2\pi R^2+2\pi Rh=24\pi\) \(\left(cm^2\right)\)

Bài 1: 

a: ĐKXĐ: \(x\ge-\dfrac{1}{3}\)

b: Ta có: \(2\sqrt{3}=\sqrt{12}\)

\(3\sqrt{2}=\sqrt{18}\)

mà 12<18

nên \(2\sqrt{3}< 3\sqrt{2}\)

c: Ta có: \(\dfrac{12}{\sqrt{5}-2}=12\sqrt{5}+24\)

d: Ta có: \(\dfrac{24}{2-x}\cdot\sqrt{\dfrac{x^2-4x+4}{36}}\)

\(=\dfrac{24}{2-x}\cdot\dfrac{2-x}{6}\)

=4