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e: \(=\left|3-\sqrt{2}\right|=3-\sqrt{2}\)

h: \(=3-\sqrt{2}+3+\sqrt{2}=6\)

g: \(=\left|0.1-\sqrt{0.1}\right|=0.1-\sqrt{0.1}\)

i: \(=\left|2\sqrt{2}-3\right|=3-2\sqrt{2}\)

c: \(=\left|2+5\right|=7\)

o: \(=5-2\sqrt{6}-5-2\sqrt{6}=-4\sqrt{6}\)

n: \(=4-2\sqrt{3}+4+2\sqrt{3}=8\)

m: \(=7+2\sqrt{10}-7-2\sqrt{10}=0\)

1

Có: \(tgB=\dfrac{CA}{CB}=\dfrac{0,9}{1,2}=\dfrac{3}{4}\)

\(cotgB=\dfrac{CB}{CA}=\dfrac{1,2}{0,9}=\dfrac{4}{3}\)

Vì A, B phụ nhau nên:

\(cotgA=tgB=\dfrac{3}{4}\\ tgA=cotgB=\dfrac{4}{3}\)

Áp dụng pytago vào tam giác ABC vuông tại C, có:

\(AB^2=BC^2+AC^2=1,2^2+0,9^2=1,5^2\Rightarrow AB=1,5\left(vì.AB>0\right)\)

Do đó: \(sinB=\dfrac{CA}{AB}=\dfrac{0,9}{1,5}=\dfrac{3}{5};cosB=\dfrac{CB}{BA}=\dfrac{1,2}{1,5}=\dfrac{4}{5}\)

Vì A, B phụ nhau nên:

\(sinA=cosB=\dfrac{4}{5};cosA=sinB=\dfrac{3}{5}\)

3:

a: Xét ΔBAC có AB^2=CA^2+CB^2

nên ΔABC vuông tại C

b: sin A=cos B=BC/AC=căn 15/5

cos A=sin A=CA/BC=căn 2/5=1/5*căn 10

tan A=cot B=căn 15/căn 10=căn 3/2

cot A=tan B=căn 2/3

Bài 1:

\(a,A=6\sqrt{2}-6\sqrt{2}+2\sqrt{5}=2\sqrt{5}\\ b,B=\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}+\dfrac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}=\sqrt{3}+\sqrt{2}\\ c,=2\sqrt{3}-6\sqrt{3}+15\sqrt{3}-4\sqrt{3}=7\sqrt{3}\\ d,=1+6\sqrt{3}-\sqrt{3}-1=5\sqrt{3}\\ e,=4\sqrt{2}+\sqrt{2}-6\sqrt{2}+3\sqrt{2}=2\sqrt{2}\)

Bài 2:

\(a,ĐK:x\ge\dfrac{3}{2}\\ PT\Leftrightarrow\sqrt{2x-3}=5\Leftrightarrow2x-3=25\Leftrightarrow x=14\\ b,PT\Leftrightarrow x^2=\sqrt{\dfrac{98}{2}}=\sqrt{49}=7\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=-\sqrt{7}\end{matrix}\right.\\ c,ĐK:x\ge3\\ PT\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}+1\right)=0\\ \Leftrightarrow\sqrt{x-3}=0\left(\sqrt{x+3}+1>0\right)\\ \Leftrightarrow x=3\\ d,ĐK:x\ge1\\ PT\Leftrightarrow2\sqrt{x-1}-\sqrt{x-1}+3\sqrt{x-1}=4\\ \Leftrightarrow\sqrt{x-1}=1\Leftrightarrow x=2\left(tm\right)\\ e,PT\Leftrightarrow2x-1=16\Leftrightarrow x=\dfrac{17}{2}\\ f,PT\Leftrightarrow\left|2x-1\right|=\sqrt{3}-1\Leftrightarrow\left[{}\begin{matrix}2x-1=\sqrt{3}-1\\2x-1=1-\sqrt{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{3}}{2}\\x=\dfrac{2-\sqrt{3}}{2}\end{matrix}\right.\)

Bài 3:

\(a,Q=\dfrac{1+5}{3-1}=3\\ b,P=\dfrac{x+\sqrt{x}-6+x-2\sqrt{x}-3-x+4\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\ P=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-3}\\ c,M=\dfrac{\sqrt{x}}{\sqrt{x}-3}\cdot\dfrac{3-\sqrt{x}}{\sqrt{x}+5}=\dfrac{-\sqrt{x}}{\sqrt{x}+5}\)

Vì \(-\sqrt{x}\le0;\sqrt{x}+5>0\) nên \(M< 0\)

Do đó \(\left|M\right|>\dfrac{1}{2}\Leftrightarrow M< -\dfrac{1}{2}\Leftrightarrow-\dfrac{\sqrt{x}}{\sqrt{x}+5}+\dfrac{1}{2}< 0\)

\(\Leftrightarrow\dfrac{2\sqrt{x}-\sqrt{x}-5}{2\left(\sqrt{x}+5\right)}< 0\Leftrightarrow\sqrt{x}-5< 0\left(\sqrt{x}+5>0\right)\\ \Leftrightarrow0\le x< 25\)

Bài 4:

\(a,A=\dfrac{16+2\cdot4+5}{4-3}=29\\ b,B=\dfrac{2\sqrt{x}-9-x+9+2x-3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ B=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\\ c,P=\dfrac{x+2\sqrt{x}+5}{\sqrt{x}-3}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{x+2\sqrt{x}+5}{\sqrt{x}+1}\\ P=\dfrac{\left(\sqrt{x}+1\right)^2+4}{\sqrt{x}+1}=\sqrt{x}+1+\dfrac{4}{\sqrt{x}+1}\\ P\ge2\sqrt{\left(\sqrt{x}+1\right)\cdot\dfrac{4}{\sqrt{x}+1}}=2\sqrt{4}=4\\ P_{min}=4\Leftrightarrow\left(\sqrt{x}+1\right)^2=4\Leftrightarrow\sqrt{x}+1=2\Leftrightarrow x=1\left(tm\right)\)

a.

Với \(m=-1\) pt trở thành: \(x^2+4x-2=0\)

\(\Delta'=4+2=6>0\) nên pt có 2 nghiệm pb:

\(x_1=-2+\sqrt{6}\) ; \(x_2=-2-\sqrt{6}\)

b.

\(\Delta'=\left(m-1\right)^2-\left(m^2-3\right)=-2m+4\ge0\Rightarrow m\le2\)

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m-1\right)\\x_1x_2=m^2-3\end{matrix}\right.\)

\(x_1\left(x_1-x_2\right)+x_2^2=33\)

\(\Leftrightarrow x_1^2+x_2^2-x_1x_2=33\)

\(\Leftrightarrow\left(x_1+x_2\right)^2-3x_1x_2=33\)

\(\Leftrightarrow4\left(m-1\right)^2-3\left(m^2-3\right)=33\)

\(\Leftrightarrow m^2-8m-20=0\Rightarrow\left[{}\begin{matrix}m=10>2\left(loại\right)\\m=-2\end{matrix}\right.\)

\(\left(x+2\right)\left(\dfrac{360}{x}-6\right)=360\)

\(ĐK:x\ne0\)

\(\Leftrightarrow\left(x+2\right)\left(\dfrac{360-6x}{x}\right)=360\)

\(\Leftrightarrow360-6x+\dfrac{720-12x}{x}=360\)

\(\Leftrightarrow360x-6x^2+720-12x=360x\)

\(\Leftrightarrow6x^2+12x-720=0\)

\(\Delta=12^2-4.6.\left(-720\right)\)

    \(=17424>0\)

`->` pt có 2 nghiệm

\(\left\{{}\begin{matrix}x_1=\dfrac{-12-\sqrt{17424}}{12}=-12\\x_2=\dfrac{-12+\sqrt{17424}}{12}=10\end{matrix}\right.\) ( tm )

Vậy \(S=\left\{-12;10\right\}\)

a.

Với \(m=3\) pt trở thành: \(2x^2+5x+2=0\)

\(\Delta=5^2-4.2.2=9>0\) nên pt có 2 nghiệm phân biệt:

\(x_1=\dfrac{-5+\sqrt{9}}{2.2}=-\dfrac{1}{2}\)

\(x_2=\dfrac{-5-\sqrt{9}}{2.2}=-2\)

b.

\(\Delta=\left(2m-1\right)^2-8\left(m-1\right)=4m^2-12m+9=\left(2m-3\right)^2\ge0;\forall m\)

Phương trình luôn có 2 nghiệm với mọi m

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{2m-1}{2}\\x_1x_2=\dfrac{m-1}{2}\end{matrix}\right.\)

\(4x_1^2+2x_1x_2+4x_2^2=1\)

\(\Leftrightarrow4\left(x_1^2+2x_1x_2+x_2^2\right)-6x_1x_2=1\)

\(\Leftrightarrow4\left(x_1+x_2\right)^2-6x_1x_2=1\)

\(\Leftrightarrow\left(2m-1\right)^2-3\left(m-1\right)=1\)

\(\Leftrightarrow4m^2-7m+3=0\Rightarrow\left[{}\begin{matrix}m=1\\m=\dfrac{3}{4}\end{matrix}\right.\)