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\(Q=x-2-2\sqrt{x-2}+4\)
\(=\left(\sqrt{x-2}-1\right)^2+3>=3\)
Dấu = xảy ra khi x=3
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow x=0\Leftrightarrow\left\{{}\begin{matrix}5-m=y\\3+m=y\end{matrix}\right.\Leftrightarrow y=4\Leftrightarrow5-m=4\Leftrightarrow m=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`a,` ĐKXĐ: `x>=0;x\ne1`
`A=...=(sqrtx(1+sqrtx)+sqrtx(1-sqrtx)+sqrtx-3)/((1-sqrtx)(1+sqrtx))`
`=(sqrtx+x+sqrtx-x+sqrtx-3)/((1-sqrtx)(1+sqrtx))`
`=(3sqrtx-3)/((1-sqrtx)(1+sqrtx))`
`=-3/(1+sqrtx)`
`b,A=-3/(1+sqrtx)`
Vì `x>=0` nên `1+sqrtx>=1` nên `3/(1+sqrtx)<=3` suy ra `A>=-3`
Dấu "=" xảy ra `<=>x=0`
Vậy `A_(min)=-3<=>x=0`
![](https://rs.olm.vn/images/avt/0.png?1311)
b: Xét ΔABH vuông tại H có
\(AB^2=AH^2+HB^2\)
hay AH=12(cm)
Xét ΔAHB vuông tại H có
\(\sin\widehat{B}=\cos\widehat{C}=\dfrac{AH}{AB}=\dfrac{12}{13}\)
\(\cos\widehat{B}=\sin\widehat{C}=\dfrac{5}{13}\)
\(\tan\widehat{B}=\cot\widehat{C}=\dfrac{12}{5}\)
\(\cot\widehat{B}=\tan\widehat{C}=\dfrac{5}{12}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 2:
Ta có: \(x^2-2\left(m+1\right)x+m^2+4=0\)
a=1; b=-2m-2; \(c=m^2+4\)
\(\text{Δ}=b^2-4ac\)
\(=\left(-2m-2\right)^2-4\cdot\left(m^2+4\right)\)
\(=4m^2+8m+4-4m^2-16\)
=8m-12
Để phương trình có hai nghiệm phân biệt thì Δ>0
\(\Leftrightarrow8m>12\)
hay \(m>\dfrac{3}{2}\)
Áp dụng hệ thức Vi-et, ta được:
\(\left\{{}\begin{matrix}x_1+x_2=2\left(m+1\right)=2m+2\\x_1x_2=m^2+4\end{matrix}\right.\)
Vì x1 là nghiệm của phương trình nên ta có:
\(x_1^2-2\left(m+1\right)\cdot x_1+m^2+4=0\)
\(\Leftrightarrow x_1^2=2\left(m+1\right)x_1-m^2-4\)
Ta có: \(x_1^2+2\left(m+1\right)x_2=2m^2+20\)
\(\Leftrightarrow2\left(m+1\right)x_1-m^2-4+2\left(m+1\right)x_2-2m^2-20=0\)
\(\Leftrightarrow2\left(m+1\right)\left(x_1+x_2\right)-3m^2-24=0\)
\(\Leftrightarrow2\left(m+1\right)\cdot\left(2m+2\right)-3m^2-24=0\)
\(\Leftrightarrow4m^2+8m+4-3m^2-24=0\)
\(\Leftrightarrow m^2+8m-20=0\)
Đến đây bạn tự tìm m là xong rồi
![](https://rs.olm.vn/images/avt/0.png?1311)
1. ĐKXĐ: $x\geq 1$
PT $\Leftrightarrow \sqrt{x-1}=13-x$
\(\Rightarrow \left\{\begin{matrix} 13-x\geq 0\\ x-1=(13-x)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\leq 13\\ x^2-27x+170=0\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x\leq 13\\ (x-17)(x-10)=0\end{matrix}\right.\)
\(\Rightarrow x=10\) (tm)
2. ĐKXĐ: $x\geq 3$
\(3\sqrt{x+34}-3\sqrt{x-3}=1\)
\(\Leftrightarrow 3\sqrt{x+34}=3\sqrt{x-3}+1\)
\(\Rightarrow 9(x+34)=9x+6\sqrt{x-3}-26\)
\(\Leftrightarrow \frac{166}{3}=\sqrt{x-3}\)
$\Leftrightarrow x-3=\frac{27556}{9}$
$\Leftrightarrow x=\frac{27583}{9}$ (tm)
![](https://rs.olm.vn/images/avt/0.png?1311)
Xét (I) có
\(\widehat{BAC}=\dfrac{1}{2}\widehat{BIC}\)
nên \(\widehat{BIC}=2\cdot31^0=62^0\)
Xét (K) có
\(\widehat{DIE}=\dfrac{1}{2}\cdot\widehat{DKE}\)
nên \(\widehat{DKE}=2\cdot62^0=124^0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\widehat{BAC}=\dfrac{1}{2}sđ\stackrel\frown{BC}\)
\(\widehat{CBx}=\dfrac{1}{2}sđ\stackrel\frown{BC}\)
Do đó: \(\widehat{BAC}=\widehat{CBx}\)
mà \(\widehat{BOC}=2\cdot\widehat{BAC}\)
nên \(\widehat{BOC}=2\cdot\widehat{CBx}\)
\(\sqrt{x^2-2x\sqrt{11}+11}=10\)
\(< =>\sqrt{\left(x-\sqrt{11}\right)^2}=10\\ < =>\left|x-\sqrt{11}\right|=10\\ < =>\left[{}\begin{matrix}x-\sqrt{11}=10\\x-\sqrt{11}=-10\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=10+\sqrt{11}\\x=-10+\sqrt{11}\end{matrix}\right.\)
\(1-\sqrt{1+5x}=x\left(đk:x\ge-\dfrac{1}{5}\right)\\ < =>\sqrt{1+5x}=1-x\\ < =>\left\{{}\begin{matrix}1-x\ge0\\1+5x=1-2x+x^2\end{matrix}\right.\\ < =>\left\{{}\begin{matrix}x\le1\\x^2-7x=0\end{matrix}\right.\\ < =>\left\{{}\begin{matrix}x\le1\\x\left(x-7\right)=0\end{matrix}\right.\\ < =>\left\{{}\begin{matrix}x\le1\\\left[{}\begin{matrix}x=0\left(tm\right)\\x=7\left(ktm\right)\end{matrix}\right.\end{matrix}\right.\\ < =>x=0\)