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\(ĐK:x\ge-1\)
Đặt \(\left\{{}\begin{matrix}a=\sqrt{x+1}\\b=\sqrt{x^2-x+1}\\c=\sqrt{x+3}\end{matrix}\right.\left(a,b\ge0;c>0\right)\), pt trở thành:
\(\dfrac{ab}{c}+a=b+c\\ \Leftrightarrow ab+ac=bc+c^2\\ \Leftrightarrow ab+ac-bc-c^2=0\\ \Leftrightarrow\left(b+c\right)\left(a-c\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a=c\\b+c=0\end{matrix}\right.\\ \forall a=c\Leftrightarrow x+1=x+3\Leftrightarrow0x=1\left(vn\right)\\ \forall b+c=0\)
Vì \(b\ge0;c>0\Leftrightarrow b+c>0\) nên pt vô nghiệm
Vậy pt vô nghiệm
a),b) Áp dụng tslg trong tam giác ABC vuông tại A:
\(\left\{{}\begin{matrix}sinB=\dfrac{AC}{BC}=\dfrac{12}{13}\\sinC=\dfrac{AB}{BC}=\dfrac{5}{13}\end{matrix}\right.\)
c) Ta có: \(sinB=\dfrac{12}{13}\Rightarrow\widehat{B}\approx67^0\)
\(sinC=\dfrac{5}{13}\Rightarrow\widehat{C}\approx23^0\)
\(=\sqrt{\left(\sqrt{6}+\sqrt{2}\right)^2}+\dfrac{14\left(3-\sqrt{2}\right)}{9-2}-\dfrac{\sqrt{3}\left(\sqrt{6}-\sqrt{2}\right)}{\sqrt{3}}\\ =\sqrt{6}+\sqrt{2}+2\left(3-\sqrt{2}\right)-\sqrt{6}+\sqrt{2}\\ =2\sqrt{2}+6-2\sqrt{2}=6\)
\(\sqrt{8+4\sqrt{3}}+\dfrac{14}{3+\sqrt{2}}-\dfrac{3\sqrt{2}-\sqrt{6}}{\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{6}+\sqrt{2}\right)^2}+\dfrac{14\left(3-\sqrt{2}\right)}{9-2}-\dfrac{\sqrt{3}\left(\sqrt{6}-\sqrt{2}\right)}{\sqrt{3}}\)
\(=\sqrt{6}+\sqrt{2}+6-2\sqrt{2}-\sqrt{6}+\sqrt{2}=6\)
Lời giải:
Áp dụng HTL trong tam giác vuông:
$AB^2=BH.BC=BH(BH+CH)$
$\Leftrightarrow 3^2=x(x+3,2)$
$\Leftrightarrow x^2+3,2x-9=0$
$\Leftrightarrow (x-1,8)(x+5)=0$
$\Rightarrow x=1,8$ (do $x>0$)
Vậy $x=1,8$ (cm)
Bài 1:
Áp dụng HTL trong tam giác vuông:
$AB^2=BH.BC$
$\Rightarrow BH=\frac{AB^2}{BC}=\frac{6^2}{10}=3,6$ (cm)
$CH=BC-BH=10-3,6=6,4$ (cm)
Tiếp tục áp dụng HTL:
$AH^2=BH.CH=3,6.6,4$
$\Rightarrow AH=4,8$ (cm)
$AC^2=CH.BC=6,4.10=64$
$\Rightarrow AC=8$ (cm)
Bài 2:
Áp dụng định lý Pitago:
$BC=\sqrt{AB^2+AC^2}=\sqrt{3^2+1^2}=2$ (cm)
$AH=\frac{2S_{ABC}}{BC}=\frac{AB.AC}{BC}=\frac{\sqrt{3}.1}{2}=\frac{\sqrt{3}}{2}$ (cm)
$BH=\sqrt{AB^2-AH^2}=\sqrt{3-\frac{3}{4}}=\frac{3}{2}$ (cm)
$CH=BC-BH=2-\frac{3}{2}=\frac{1}{2}$ (cm)
\(1,\\ a,=\sqrt{\left(3+\sqrt{7}\right)^2}-\sqrt{\left(\sqrt{7}-1\right)^2}=3+\sqrt{7}-\sqrt{7}+1=4\\ b,K=\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}\left(\sqrt{3}-1\right)}=\dfrac{\sqrt{3}-1}{\sqrt{2}\left(\sqrt{3}-1\right)}=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\\ c,=\sqrt{\left(6-2\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-4\right)^2}=6-2\sqrt{6}+2\sqrt{6}-4=2\\ e,=\sqrt{\left(2-\sqrt{2}\right)^2}-\left(\sqrt{6}-\sqrt{2}\right)=2-\sqrt{2}-\sqrt{6}+\sqrt{2}=2-\sqrt{6}\)
\(2,\\ a,A=\dfrac{x-3\sqrt{x}+3\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}+3}{x+9}\\ A=\dfrac{x+9}{\left(\sqrt{x}-3\right)\left(x+9\right)}=\dfrac{1}{\sqrt{x}-3}\\ b,x=4+2\sqrt{3}\Leftrightarrow\sqrt{x}=\sqrt{3}+1\\ \Leftrightarrow A=\dfrac{1}{\sqrt{3}+1-3}=\dfrac{1}{\sqrt{3}+2}=2-\sqrt{3}\)
\(a,ĐK:x\ge0\\ PT\Leftrightarrow3\sqrt{x}+\dfrac{\sqrt{x}+1}{4}=2\\ \Leftrightarrow\dfrac{13\sqrt{x}+1}{4}=2\Leftrightarrow13\sqrt{x}+1=8\\ \Leftrightarrow\sqrt{x}=\dfrac{7}{13}\Leftrightarrow x=\dfrac{49}{169}\\ b,ĐK:x\ge0\\ PT\Leftrightarrow4\sqrt{2x}-2\sqrt{2x}+\dfrac{1}{4}\cdot4\sqrt{2x}=0\\ \Leftrightarrow3\sqrt{2x}=0\\ \Leftrightarrow2x=0\Leftrightarrow x=0\left(tm\right)\)
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