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a) ta có : \(A=\dfrac{sin33}{cos57}+\dfrac{tan32}{cot58}-2\left(sin20.cos70+cos20.sin70\right)\)
\(\Leftrightarrow A=\dfrac{sin33}{cos\left(90-33\right)}+\dfrac{tan32}{cot\left(90-32\right)}-2\left(sin20.cos\left(90-20\right)+cos20.sin\left(90-20\right)\right)\)
\(\Leftrightarrow A=\dfrac{sin33}{sin33}+\dfrac{tan32}{tan32}-2\left(sin20.sin20+cos20.cos20\right)\)\(\Leftrightarrow A=1+1-2\left(sin^220+cos^220\right)=1+1-2=0\)
b) sữa đề chút nha
ta có : \(B=\dfrac{sin^215+sin^275-sin^212-sin^278}{cos^213+cos^277+cos^21+cos^289}+\dfrac{2tan55}{cot35}\)
\(\Leftrightarrow B=\dfrac{sin^215+sin^2\left(90-15\right)-sin^212-sin^2\left(90-12\right)}{cos^213+cos^2\left(90-13\right)+cos^21+cos^2\left(90-1\right)}+\dfrac{2tan\left(90-35\right)}{cot35}\)
\(\Leftrightarrow B=\dfrac{sin^215+cos^215-sin^212-cos^212}{cos^213+sin^213+cos^21+sin^21}+\dfrac{2cot35}{cot35}\) \(\Leftrightarrow B=\dfrac{sin^215+cos^215-\left(sin^212+cos^212\right)}{cos^213+sin^213+cos^21+sin^21}+\dfrac{2cot35}{cot35}\)\(\Leftrightarrow B=\dfrac{1-1}{cos^213+sin^213+cos^21+sin^21}+2=0+2=2\)
\(a,A=\dfrac{3x+2-3x+2+3x-6}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{3x-2}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{1}{3x+2}\\ b,B=\dfrac{1}{2}+\dfrac{x}{\dfrac{x+2-x}{x+2}}=\dfrac{1}{2}+\dfrac{x}{\dfrac{2}{x+2}}=\dfrac{1}{2}+\dfrac{x\left(x+2\right)}{2}\\ B=\dfrac{1+x^2+2x}{2}=\dfrac{\left(x+1\right)^2}{2}\)
`a)đk:a>0,a ne 9`
`A=((sqrta+3+sqrta-3)/(a-9)).((sqrta-3)/sqrta)`
`=((2sqrtx)/(a-9)).((sqrta-3)/sqrta)`
`=2/(sqrta+3)`
`b)A>1/2`
`<=>2/(sqrta+3)>1/2`
`<=>sqrta+3<4`
`<=>sqrta<1`
`<=>a<1`
KẾt hợp đkxđ:`0<x<1`
ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\a\ne9\end{matrix}\right.\)
a) Ta có: \(A=\left(\dfrac{1}{\sqrt{a}-3}+\dfrac{1}{\sqrt{a}+3}\right)\left(1-\dfrac{3}{\sqrt{a}}\right)\)
\(=\dfrac{\sqrt{a}+3+\sqrt{a}-3}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\cdot\dfrac{\sqrt{a}-3}{\sqrt{a}}\)
\(=\dfrac{2\sqrt{a}}{\sqrt{a}+3}\cdot\dfrac{1}{\sqrt{a}}\)
\(=\dfrac{2}{\sqrt{a}+3}\)
b) Để \(A>\dfrac{1}{2}\) thì \(A-\dfrac{1}{2}>0\)
\(\Leftrightarrow\dfrac{2}{\sqrt{a}+3}-\dfrac{1}{2}>0\)
\(\Leftrightarrow\dfrac{4-\left(\sqrt{a}+3\right)}{2\left(\sqrt{a}+3\right)}>0\)
mà \(2\left(\sqrt{a}+3\right)>0\forall a\)
nên \(1-\sqrt{a}>0\)
\(\Leftrightarrow\sqrt{a}< 1\)
hay a<1
Kết hợp ĐKXĐ, ta được: 0<a<1
a: ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
Ta có: \(B=\dfrac{6-7x}{x^2-4}+\dfrac{3}{x+2}-\dfrac{2}{2-x}\)
\(=\dfrac{6-7x+3x-6+2x+4}{\left(x+2\right)\left(x-2\right)}\)
\(=\dfrac{-2x+4}{\left(x+2\right)\left(x-2\right)}\)
\(=-\dfrac{2}{x+2}\)
\(a,ĐK:x\ne\pm1;x\ne0\\ M=\dfrac{1-x+2x}{\left(1+x\right)\left(1-x\right)}:\dfrac{1-x}{x}\\ M=\dfrac{x+1}{\left(x+1\right)\left(1-x\right)}\cdot\dfrac{x}{1-x}=\dfrac{x}{\left(1-x\right)^2}\\ b,ĐK:x\ge0;x\ne4\\ N=\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ N=\dfrac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)
Tất cả đều phải tìm điều kiện
\(\dfrac{2}{x^2-y^2}.\sqrt{\dfrac{3x^2+6xy+3y^2}{4}}\)
\(ĐK:x\ne\pm y\)
\(=\dfrac{2\left|x+y\right|}{2\left(x+y\right)\left(x-y\right)}=\dfrac{\sqrt{3}\left|x+y\right|}{\left(x+y\right)\left(x-y\right)}\)
Nếu x > -y thì x + y > 0 , ta có :\(\dfrac{\sqrt{3}}{x-y}\)
Nếu x < -y thì x + y < 0 , ta có :\(\dfrac{-\sqrt{3}}{x-y}\)
\(\dfrac{1}{\sqrt{3}+1}+\dfrac{1}{\sqrt{3}-1}-2\sqrt{3}\)
\(=\dfrac{\sqrt{3}-1}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}+\dfrac{\sqrt{3}+1}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}-2\sqrt{3}\)
\(=\dfrac{\sqrt{3}-1+\sqrt{3}+1}{3-1}-2\sqrt{3}\)
\(=\dfrac{2\sqrt{3}}{2}-2\sqrt{3}\)
\(=\sqrt{3}-2\sqrt{3}=-\sqrt{3}\)
\(A=\dfrac{x+12}{x-4}+\dfrac{1}{\sqrt{x}+2}-\dfrac{4}{\sqrt{x}-2}\)
\(=\dfrac{x+12+\sqrt{x}-2-4\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x+\sqrt{x}+10-4\sqrt{x}-8}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x-3\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)
\(=\dfrac{\sin35}{\cos35}\cdot\cot35+\dfrac{\cos55}{\sin55}\cdot\cot35\)
\(=\cot35\cdot\left(\dfrac{\sin35}{\cos35}+\dfrac{\sin35}{\cos35}\right)\)
\(=\dfrac{\cos35}{\sin35}\cdot\dfrac{2\sin35}{\cos35}=2\)