ko ghi lại đề

\(8x^2+8x+6=\left(5x+4\right)\sqrt{x^2+3}\)\(3\)

bình hai vế ta đc

\(64x^2+64x+36=\left(5x+4\right)^2x^2+3\)

\(64.\left(x^2+x\right)+36=25x+16x^2+3\)

\(64.\left(x^2+x\right)+36=16\left(x+x^2\right)+9+3\)

\(64\left(x^2+x\right)+36=16\left(x+x^2\right)+12\)

\(=64-\left(x^2+x\right)+36-16\left(x+x^2\right)-12\)

\(=72\)

bài nay ko cần điều kiện

\(\hept{\begin{cases}x+2y=5\left(1\right)\\\sqrt{2}x+y=4\left(2\right)\end{cases}\Leftrightarrow\hept{\begin{cases}x+2y=5\\2\sqrt{2}x+2y=8\end{cases}}}\)

Tru ve voi ve cua (1) va (2) ta duoc:

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

\(\Leftrightarrow x=\frac{3}{2\sqrt{2}-1}\)

\(y=\frac{5\sqrt{2}-4}{2\sqrt{2}-1}\)

Vay nghiem cua HPT la \(\left(\frac{3}{2\sqrt{2}-1};\frac{5\sqrt{2}-4}{2\sqrt{2}-1}\right)\)

1.\(DK:x\le\frac{1}{3}\)

2.\(DK:x\ge-1\)

3.\(DK:-1\le x< 1\)

a.\(1-\sin^2\alpha=\cos^2\alpha\)

b.\(\sin^4\alpha+\cos^4\alpha+2\sin^2\alpha.\cos^2\alpha=\left(\sin^2\alpha+\cos^2\alpha\right)^2=1\)

c.\(\left(1-\cos\alpha\right)\left(1+\cos\alpha\right)=1-\cos^2\alpha=\sin^2\alpha\)

d.\(1+\sin^2\alpha+\cos^2\alpha=1+1=2\)

e.\(\tan^2\alpha-\sin^2\alpha.\tan^2\alpha=\tan^2\alpha\left(1-\sin^2\alpha\right)=\tan^2\alpha.\cos^2\alpha=\sin^2\alpha\)

g.\(\cos^2\alpha+\cos^2\alpha.\tan^2\alpha=\cos^2\alpha\left(1+\tan^2\alpha\right)=\cos^2\alpha.\frac{1}{\cos^2\alpha}=1\)

a) A có nghĩa\(\Leftrightarrow x-y\ne0\Leftrightarrow x\ne y\)

b) \(A=\frac{x+y-2\sqrt{xy}}{x-y}=\frac{\left(\sqrt{x-\sqrt{y}}\right)^2}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}=\frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)