ΔCAB vuông tại C có CH là đường cao

nên AC^2=AH*AB

=>AH*(AH+16)=15^2=225

=>AH=9cm

=>BA=9+16=25cm

BC=căn 25^2-15^2=20cm

S ABC=1/2*15*20=150cm²

\(3\sqrt{x-5}=6\) (ĐK: \(x\ge5\))

\(\Leftrightarrow\sqrt{x-5}=\dfrac{6}{3}\)

\(\Leftrightarrow\sqrt{x-5}=2\)

\(\Leftrightarrow x-5=2^2\)

\(\Leftrightarrow x-5=4\)

\(\Leftrightarrow x=4+5\)

\(\Leftrightarrow x=9\left(tm\right)\)

\(\left(x+\dfrac{1}{x}\right)^2-4\left(x+\dfrac{1}{x}\right)^2+3=0\\\Leftrightarrow3\left(x+\dfrac{1}{x}\right)^2=3\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{x}=1\\x+\dfrac{1}{x}=-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2-x+1=0\\x^2+x+1=0\end{matrix}\right.\\ \Leftrightarrow x\in\varnothing \)

Lời giải:
ĐKXĐ: $9x^2+6x+1\geq 0$

$\Leftrightarrow (3x+1)^2\geq 0$

$\Leftrightarrow x\in\mathbb{R}$
--------------------------

$\sqrt{9x^2+6x+1}=2-x$

\(\Rightarrow \left\{\begin{matrix} 2-x\geq 0\\ 9x^2+6x+1=(2-x)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\leq 2\\ 9x^2+6x+1=x^2-4x+4\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\leq 2\\ 8x^2+10x-3=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\leq 2\\ (4x-1)(2x+3)=0\end{matrix}\right.\Leftrightarrow x=\frac{1}{4}\) hoặc $x=\frac{-3}{2}$

\(\dfrac{15}{\sqrt{6}-1}+\dfrac{8}{\sqrt{6}+2}+\dfrac{6}{3-\sqrt{6}}-9\sqrt{6}\)

\(=\dfrac{15\left(\sqrt{6}+1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}+\dfrac{8\left(\sqrt{6}-2\right)}{\left(\sqrt{6}+2\right)\left(\sqrt{6}-2\right)}+\dfrac{6\left(3+\sqrt{6}\right)}{\left(3-\sqrt{6}\right)\left(3+\sqrt{6}\right)}-9\sqrt{6}\)

\(=\dfrac{15\left(\sqrt{6}+1\right)}{6-1}+\dfrac{8\left(\sqrt{6}-2\right)}{6-4}+\dfrac{6\left(3+\sqrt{6}\right)}{9-6}-9\sqrt{6}\)

\(=3\left(\sqrt{6}+1\right)+4\left(\sqrt{6}-2\right)+2\left(3+\sqrt{6}\right)-9\sqrt{6}\)

\(=3\sqrt{6}+3+4\sqrt{6}-8+6+2\sqrt{6}-9\sqrt{6}\)

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

\(=1\)

\(\sqrt{\left(\sqrt{5}-1\right)\sqrt{14-6\sqrt{5}}}\)

\(=\sqrt{\left(\sqrt{5}-1\right)\sqrt{9-6\sqrt{5}+5}}\)

\(=\sqrt{\left(\sqrt{5}-1\right)\sqrt{3^2-2\cdot3\cdot\sqrt{5}+\left(\sqrt{5}\right)^2}}\)

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

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

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

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

\(=\sqrt{4\sqrt{5}-8}\)

\(=\sqrt{4\left(\sqrt{5}-2\right)}\)

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

mình cần gấp ạ

Bài 1: 

a) \(A=\sqrt{\left(2-\sqrt{3}\right)^2}+2\sqrt{3}\)

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

\(A=2-\sqrt{3}+2\sqrt{3}\)

\(A=2+\sqrt{3}\)

b) \(B=\sqrt{18}-2\sqrt{50}+3\sqrt{8}+\sqrt[3]{27}\)

\(B=3\sqrt{2}-10\sqrt{2}+6\sqrt{2}+3\)

\(B=\left(3-10+6\right)\sqrt{2}+3\)

\(B=-\sqrt{2}+3\)

c) \(C=\dfrac{4}{\sqrt{5}-1}-\dfrac{10}{\sqrt{5}}+\dfrac{\sqrt{125}}{\sqrt{5}}+\sqrt{2}\sqrt{\dfrac{5}{2}}\)

\(C=1+\sqrt{5}-2\sqrt{5}+\dfrac{5\sqrt{5}}{\sqrt{5}}+\sqrt{\dfrac{2\cdot5}{2}}\)

\(C=1+\sqrt{5}-2\sqrt{5}+5+\sqrt{5}\)

\(C=6\)

a: y=150000000*(1+6,8%)*x=160200000x

b: Sau 5 năm số tiền nhận được là:

160200000*5=801000000(đồng)

Sau 5 năm số tiền nhận được là:801000000

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

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

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

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

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

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

\(P=2-1\)

\(P=1\)

a: ĐKXĐ: x-1>=0 và x-2<>0

=>x>=1 và x<>2

b: ĐKXĐ: 3-x>0 và 5-2x>=0 và x<>0

=>x<3 và x<>0 và x<=5/2

=>x<=5/2 và x<>0

c: ĐKXĐ: (x-3)(x-5)>=0

=>x>=5 hoặc x<=3

d: ĐKXĐ: x-1>0 và x<>0 và x-3>=0

=>x>=3

\(a,\dfrac{\sqrt{x-1}+x}{x-2}\) xác định \(\Leftrightarrow\left[{}\begin{matrix}x-1\ge0\\x-2\ne0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x\ge1\\x\ne2\end{matrix}\right.\)

\(b,\dfrac{x-2}{\sqrt{3-x}}+\dfrac{\sqrt{5-2x}}{x}\) xác định \(\Leftrightarrow\left[{}\begin{matrix}3-x>0\\x\ne0\\5-2x\ge0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x< 3\\x\ne0\\x\le\dfrac{5}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x\ne0\\x\le\dfrac{5}{2}\end{matrix}\right.\)

\(c,\sqrt{x^2-8x+15}\) xác định \(\Leftrightarrow x^2-8x+15\ge0\Leftrightarrow x^2-5x-3x+15\ge0\)

\(\Leftrightarrow x\left(x-5\right)-3\left(x-5\right)\ge0\\\Leftrightarrow\left(x-3\right)\left(x-5\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x-3\ge0\\x-5\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x-3\le0\\x-5\le0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\ge3\\x\ge5\end{matrix}\right.\\\left[{}\begin{matrix}x\le3\\x\le5\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x\ge5\\x\le3\end{matrix}\right.\)

\(d,\dfrac{3}{x}+\sqrt{\dfrac{2}{x-1}}-\sqrt{\dfrac{x-3}{3}}\) xác định \(\Leftrightarrow\left[{}\begin{matrix}x\ne0\\x-1\ge0\\x-3\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x\ne0\\x\ge1\\x\ge3\end{matrix}\right.\) \(\Leftrightarrow x\ge3\)

a: BC=căn 6^2+8^2=10cm

BH=AB^2/BC=3,6cm

CH=10-3,6=6,4cm

sin ABC=AC/BC=4/5

=>góc ABC=53 độ

b: ΔAHB vuông tại H có HE là đường cao

nên AE*AB=AH^2

ΔAHC vuông tại H có HF là đường cao

nên AF*AC=AH^2

=>AE*AB=AF*AC

=>AE/AC=AF/AB

=>ΔAEF đồng dạng với ΔACB

c: góc AEH=góc AFH=góc FAE=90 độ

=>AEHF là hình chữ nhật

góc KAC+góc AFE

=góc AHE+góc KCA

=góc ABC+góc ACB=90 độ

=>AK vuông góc EF