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Câu 1:
a: Để hàm số đồng biến thì m-5>0
hay m>5
b: Để hàm số nghịch biến thì m-5<0
hay m<5
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{4x}{1-x^2}=\sqrt{5}\) ĐKXĐ : x khác 1
\(\Rightarrow4x=\sqrt{5}\left(1-x^2\right)\)
\(\Leftrightarrow4x=\sqrt{5}-x^2\sqrt{5}\)
\(\Leftrightarrow x^2\sqrt{5}-4x-\sqrt{5}=0\)
\(\Leftrightarrow x^2\sqrt{5}-5x+x-\sqrt{5}=0\)
\(\Leftrightarrow x\sqrt{5}\left(x-\sqrt{5}\right)+\left(x-\sqrt{5}\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{5}\right)\left(x\sqrt{5}+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-\sqrt{5}=0\\x\sqrt{5}=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\sqrt{5}\left(tmđk\right)\\x=-\frac{1}{\sqrt{5}}=-\frac{\sqrt{5}}{5}\left(tmđk\right)\end{cases}}}\)
\(4x=\sqrt{5}-\sqrt{5}x^2\)
\(\Rightarrow4x+\sqrt{5}x^2=\sqrt{5}\)
\(\Rightarrow x\left(4+\sqrt{5}x\right)=\sqrt{5}\)
\(\Rightarrow x.\sqrt{5}\left(\frac{4}{\sqrt{5}}+x\right)=\sqrt{5}\)
\(\Rightarrow x.\left(\frac{4}{\sqrt{5}}+x\right)=1\)
Với x = 1 \(\Rightarrow\frac{4}{\sqrt{5}}+x=1\Rightarrow x=1-\frac{4}{\sqrt{5}}=\frac{5-4\sqrt{5}}{5}\)
Với x = -1\(\Rightarrow\frac{4}{\sqrt{5}}+x=-1\Rightarrow x=-1-\frac{4}{\sqrt{5}}=-\frac{5+4\sqrt{5}}{5}\)
ko có x thỏa mãn
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\(a,=2\sqrt{2a}+3\sqrt{2a}-4\sqrt{2a}=\sqrt{2a}\\ b,=3\sqrt{5}-3+\sqrt{5}=4\sqrt{5}-3\\ c,=\dfrac{\sqrt{3}-1+\sqrt{3}+1}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\dfrac{2\sqrt{3}}{2}=\sqrt{3}\\ d,=-7+5-2\cdot\dfrac{2}{3}+\dfrac{1}{3}\cdot3=-2-\dfrac{4}{3}+1=-\dfrac{7}{3}\)
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Bài 1:
\(a,=2\sqrt{3a}-5\sqrt{3a}+\dfrac{1}{2}\cdot4\sqrt{3a}=-3\sqrt{3a}+2\sqrt{3a}=-\sqrt{3a}\\ b,=\dfrac{a+b}{b^2}\cdot\dfrac{\left|a\right|b^2}{\left|a+b\right|}=\dfrac{a+b}{b^2}\cdot\dfrac{\left|a\right|b^2}{a+b}=\left|a\right|\\ c,=5\sqrt{a}-20\left|a\right|b\sqrt{a}+20a\left|b\right|\sqrt{a}-6\sqrt{a}\\ =\left(5\sqrt{a}-6\sqrt{a}\right)-\left(20ab\sqrt{a}-20ab\sqrt{a}\right)\\ =-\sqrt{a}\\ d,=\dfrac{\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)}{x+\sqrt{3}}=x-\sqrt{3}\\ e,=\dfrac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}=1+\sqrt{a}+a\)
Bài 2:
\(a,B=7\sqrt{x+2}-4\sqrt{x+2}-2\sqrt{x+2}=\sqrt{x+2}\\ b,ĐK:x\ge-2\\ PT\Leftrightarrow\sqrt{x+2}=20\Leftrightarrow x+2=400\Leftrightarrow x=398\left(tm\right)\)
Bài 3:
\(a,M=\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\dfrac{\sqrt{a}-1}{\sqrt{a}}\\ b,M=\dfrac{\sqrt{a}-1}{\sqrt{a}}=1-\dfrac{1}{\sqrt{a}}< 1\left(\dfrac{1}{\sqrt{a}}>0\right)\)
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Xét tam giác ABC vuông tại A có AH là đường cao, ta có: \(AB^2=BH.BC=1.\left(1+4\right)=5\Rightarrow AB=\sqrt{5}cm\)
Áp dụng định lí Py-ta-go vào tam giác ABH vuông tại H có:
\(AB^2=BH^2+AH^2\Rightarrow AH^2=AB^2-BH^2=5-1=4\Rightarrow AH=2cm\)
Ta có: \(BC=BH+HC=1+4=5\left(cm\right).\)
Xét tam giác ABC vuông tại A, AH là đường cao (gt).
\(\Rightarrow AB^2=BH.BC\) (Hệ thức lượng).
\(\Rightarrow AB^2=1.5=5.\\ \Rightarrow AB=\sqrt{5}\left(cm\right).\)
\(AH^2=BH.CH\) (Hệ thức lượng).
\(\Rightarrow AH^2=1.4=4.\\ \Rightarrow AH=2\left(cm\right).\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(P=-3\left(x^2+\dfrac{4}{9}y+\dfrac{64}{9}+\dfrac{4}{3}x\sqrt{y}-\dfrac{16}{3}x-\dfrac{32}{9}\sqrt{y}\right)-\dfrac{2}{3}\left(y-2y+1\right)+2022\)
\(P=-3\left(x+\dfrac{2\sqrt{y}}{3}-\dfrac{8}{3}\right)^2-\dfrac{2}{3}\left(\sqrt{y}-1\right)^2+2022\le2022\)
\(P_{max}=2022\) khi \(\left(x;y\right)=\left(2;1\right)\)
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\(W=\dfrac{1}{2}\left(\sqrt{\left(2+\sqrt{2}\right)^2}+\sqrt{\left(2-\sqrt{2}\right)^2}\right)\\ W=\dfrac{1}{2}\left(2+\sqrt{2}+2-\sqrt{2}\right)=\dfrac{1}{2}\cdot4=2\\ Y=\dfrac{1}{2}\left(\sqrt{\left(4+\sqrt{3}\right)^2}+\sqrt{\left(4-\sqrt{3}\right)^2}\right)\\ Y=\dfrac{1}{2}\left(4+\sqrt{3}+4-\sqrt{3}\right)=\dfrac{1}{2}\cdot8=4\)
Câu 3: A
Câu 4: C
Câu 1: C
Câu 2: D
Câu 5: B
A
C
C
D
B