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Lời giải:
a. ĐKXĐ: $x>0; x\neq 4$
b.
\(M=\sqrt{x}.\left[\frac{1}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}\right].\frac{x-4}{2\sqrt{x}}\)
\(=\frac{2\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}.\frac{x-4}{2}=\frac{2\sqrt{x}}{x-4}.\frac{x-4}{2}=\sqrt{x}\)
c. Để $M>3\Leftrightarrow \sqrt{x}>3\Leftrightarrow x>9$
Kết hợp đkxđ suy ra $x>9$ thì $M>3$
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,=2\sqrt{2}\left(\sqrt{5}-1\right)\sqrt{4+\sqrt{\left(\sqrt{5}-1\right)^2}}\\ =2\sqrt{2}\left(\sqrt{5}-1\right)\sqrt{4+\sqrt{5}-1}\\ =2\left(\sqrt{5}-1\right)\sqrt{6-2\sqrt{5}}\\ =2\left(\sqrt{5}-1\right)\sqrt{\left(\sqrt{5}-1\right)^2}\\ =2\left(\sqrt{5}-1\right)^2=2\left(6-2\sqrt{5}\right)=12-4\sqrt{5}\\ b,=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8-2\sqrt{15}}\\ =\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\\ =\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)^2\\ =\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\\ =32-8\sqrt{15}+8\sqrt{15}-30=2\)
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a.
Ta có: \(\sqrt{x^2+12}>\sqrt{x^2+5}\Rightarrow\sqrt{x^2+12}-\sqrt{x^2+5}>0\)
\(\Rightarrow3x-5=\sqrt{x^2+12}-\sqrt{x^2+5}>0\Rightarrow x>\dfrac{5}{3}\)
Do đó:
\(\sqrt{x^2+12}+5=3x+\sqrt{x^2+5}\)
\(\Leftrightarrow3\left(x-2\right)+\sqrt{x^2+5}-3-\left(\sqrt{x^2+12}-4\right)=0\)
\(\Leftrightarrow3\left(x-2\right)+\dfrac{x^2-4}{\sqrt{x^2+5}+3}-\dfrac{x^2-4}{\sqrt{x^2+12}+4}=0\)
\(\Leftrightarrow\left(x-2\right)\left(3+\dfrac{x+2}{\sqrt{x^2+5}+3}-\dfrac{x+2}{\sqrt{x^2+12}+4}\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[3+\left(x+2\right)\left(\dfrac{1}{\sqrt{x^2+5}+3}-\dfrac{1}{\sqrt{x^2+12}+4}\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(3+\dfrac{\left(x+2\right)\left(\sqrt{x^2+12}-\sqrt{x^2+5}+1\right)}{\left(\sqrt{x^2+5}+3\right)\left(\sqrt{x^2+12}+4\right)}\right)=0\)
\(\Leftrightarrow x-2=0\) (do \(x>\dfrac{5}{3}\) nên ngoặc phía sau luôn dương)
\(\Leftrightarrow x=2\)
b.
\(\Delta=\left(m-1\right)^2-4\left(-m^2+m-2\right)=5m^2-6m+9=5\left(m-\dfrac{3}{5}\right)^2+\dfrac{36}{5}>0\)
Phương trình đã cho luôn có 2 nghiệm với mọi m
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=m-1\\x_1x_2=-m^2+m-2\end{matrix}\right.\)
\(Q=\left(\dfrac{x_1}{x_2}\right)^2+\left(\dfrac{x_2}{x_1}\right)^2+2-2=\left(\dfrac{x_1}{x_2}+\dfrac{x_2}{x_1}\right)^2-2\)
\(=\left(\dfrac{x_1^2+x_2^2}{x_1x_2}\right)^2-2=\left[\dfrac{\left(x_1+x_2\right)^2-2x_1x_2}{x_1x_2}\right]^2-2\)
\(=\left[\dfrac{\left(m-1\right)^2-2\left(-m^2+m-2\right)}{-m^2+m-2}\right]^2-2=\left(\dfrac{3m^2-4m+5}{m^2-m+2}\right)^2-2\)
Ta có: \(\dfrac{3m^2-4m+5}{m^2-m+2}=\dfrac{3\left(m-\dfrac{2}{3}\right)^2+\dfrac{11}{3}}{\left(m-\dfrac{1}{2}\right)^2+\dfrac{7}{4}}>0\)
\(\dfrac{3m^2-4m+5}{m^2-m+2}=\dfrac{21m^2-28m+35}{7\left(m^2-m+2\right)}=\dfrac{22\left(m^2-m+2\right)-\left(m^2+6m+9\right)}{7\left(m^2-m+2\right)}=\dfrac{22}{7}-\dfrac{\left(m+3\right)^2}{7\left(m^2-m+2\right)}\le\dfrac{22}{7}\)
\(\Rightarrow0< \dfrac{3m^2-4m+5}{m^2-m+2}\le\dfrac{22}{7}\)
\(\Rightarrow Q\le\left(\dfrac{22}{7}\right)^2-2\)
Dấu "=" xảy ra khi \(m+3=0\Leftrightarrow m=-3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 12:
a: Xét ΔABC vuông tại A có AH là đường cao ứng với cạnh huyền BC
nên \(AH^2=HB\cdot HC\)
hay HC=4,5(cm)
Xét ΔABC vuông tại A có AH là đường cao ứng với cạnh huyền BC
nên \(\left\{{}\begin{matrix}AB^2=BH\cdot BC\\AC^2=CH\cdot BC\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}AB=10\left(cm\right)\\AC=7.5\left(cm\right)\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{5}{\sqrt{7}+\sqrt{2}}-\sqrt{7}\)
\(=\sqrt{7}-\sqrt{2}-\sqrt{7}\)
\(=-\sqrt{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b: \(\text{Δ}=m^2-4m-12=\left(m-6\right)\left(m+2\right)\)
Để phương trình có hai nghiệm thì (m-6)(m+2)>=0
=>m>=6 hoặc m<=-2
\(x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2=m^2-2m-6\)
\(x_1^3+x_2^3=\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)\)
\(=\left(-m\right)^3-3\cdot\left(-m\right)\cdot\left(m+3\right)\)
\(=-m^3+3m^2+9m\)
c: \(\Leftrightarrow m^2-2m-6=9\)
\(\Leftrightarrow\left(m-5\right)\left(m+3\right)=0\)
=>m=5(loại) hay m=-3(nhận)
Bài 3: Ta có:
\(1+tan^2\alpha=\dfrac{1}{cos^2\alpha}\)
\(\Rightarrow1+\left(0,8\right)^2=\dfrac{1}{cos^2\alpha}\)
\(\Rightarrow cos^2a=\dfrac{1}{1+\left(0,8\right)^2}\)
\(\Rightarrow cos^2\alpha=\dfrac{25}{41}\)
\(\Rightarrow cos\alpha=\dfrac{5\sqrt{41}}{41}\)
Mà: \(tan\alpha=\dfrac{sin\alpha}{cos\alpha}\)
\(\Rightarrow sin\alpha=tan\alpha\cdot cos\alpha\)
\(\Rightarrow sin\alpha=0,8\cdot\dfrac{5\sqrt{41}}{41}=\dfrac{4\sqrt{41}}{41}\)
Bài 4: Ta có:
\(1+tan^2\alpha=\dfrac{1}{sin^2\alpha}\)
\(\Rightarrow sin^2\alpha=\dfrac{1}{1+tan^2\alpha}\)
\(\Rightarrow sin^2\alpha=\dfrac{1}{1+3^2}=\dfrac{1}{10}\)
\(\Rightarrow sin\alpha=\dfrac{\sqrt{10}}{10}\)
Mà: \(tan\alpha=\dfrac{sin\alpha}{cos\alpha}\)
\(\Rightarrow cos\alpha=\dfrac{sin\alpha}{tan\alpha}\)
\(\Rightarrow cos\alpha=\dfrac{\dfrac{\sqrt{10}}{10}}{3}=\dfrac{\sqrt{10}}{30}\)