\(\sqrt{x^2-2x+5}=x^2-2x-1\)

đặt \(\sqrt{x^2-2x+5}=a\left(a\ge0\right)\)

pt trở thành : \(a=a^2-6\)

\(\Leftrightarrow a^2-a-6=0\)

\(\Leftrightarrow\left(a-3\right)\left(a+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=3\left(tm\right)\\a=-2\left(loai\right)\end{cases}}\)

với \(a=3\Rightarrow\sqrt{x^2-2x+5}=3\Leftrightarrow x^2-2x+5=9\)

\(\Leftrightarrow x^2-2x-4=0\)

\(\Delta=b^2-4ac=\left(-2\right)^2-4\cdot\left(-4\right)=4+16=20\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{2+\sqrt{20}}{2}=\frac{2+2\sqrt{5}}{2}=1+\sqrt{5}\\x=\frac{2-\sqrt{20}}{2}=\frac{2-2\sqrt{5}}{2}=1-\sqrt{5}\end{cases}}\)

Ta có: \(x^2-2x+5=\left(x-1\right)^2+4\ge4\forall x\in R\)

Đặt \(t=\sqrt{x^2-2x+5}\left(t\ge2\right)\)

\(pt\Leftrightarrow t=t^2-6\Leftrightarrow t^2-t-6=0\Leftrightarrow t=-2\) (loại) hoặc t=3

Với t=3 \(\Leftrightarrow\sqrt{x^2-2x+5}=3\Leftrightarrow x^2-2x-4=0\Leftrightarrow x=1\pm\sqrt{5}\)

1, \(\sqrt{x^2-4x+4}=\sqrt{4x^2-12x+9}\)

\(\Leftrightarrow\sqrt{\left(x-2\right)^2}=\sqrt{\left(2x-3\right)^2}\Leftrightarrow\left|x-2\right|=\left|2x-3\right|\)

TH1 : \(x-2=2x-3\Leftrightarrow x=1\)

TH2 : \(x-2=3-2x\Leftrightarrow3x=5\Leftrightarrow x=\frac{5}{3}\)

2, sửa đề : \(\sqrt{x+4\sqrt{x}-4}=2\)ĐK : \(x\le-2-2\sqrt{2};-2+2\sqrt{2}\le x\)

\(\Leftrightarrow x+4\sqrt{x}-4=4\Leftrightarrow x+4\sqrt{x}-8=0\Leftrightarrow x=-2+2\sqrt{3};-2-2\sqrt{3}\)

ps : mình nghĩ đề này phải là \(\sqrt{x-4\sqrt{x}+4}=2\)nhé 

3, \(\sqrt{x^2-6x+9}=2\Leftrightarrow\sqrt{\left(x-3\right)^2}=2\Leftrightarrow\left|x-3\right|=2\)

TH1 : \(x-3=2\Leftrightarrow x=5\)

TH2 : \(x-3=-2\Leftrightarrow x=1\)

4, \(\sqrt{x^2-3x+2}=\sqrt{x-1}\)ĐK : \(x\le1;x\ge2\)

\(\Leftrightarrow\sqrt{\left(x-1\right)\left(x-2\right)}-\sqrt{x-1}=0\)

\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x-2}-1\right)=0\Leftrightarrow x=1;x=3\)

5, mình sửa đề là \(\sqrt{4x^2-4x+1}=\sqrt{x^2-6x+9}\)còn nếu đề là \(\sqrt{4x^2+4x+1}\)thì vẫn nhóm ra hđt và làm tương tự được nhé 

\(\Leftrightarrow\sqrt{\left(2x-1\right)^2}=\sqrt{\left(x-3\right)^2}\Leftrightarrow\left|2x-1\right|=\left|x-3\right|\)

TH1 : \(2x-1=x-3\Leftrightarrow x=-2\)

TH2 : \(2x-1=3-x\Leftrightarrow3x=4\Leftrightarrow x=\frac{4}{3}\)

\(\left(\frac{2}{\sqrt{3}-1}+\frac{3}{\sqrt{3}-2}+\frac{15}{3-\sqrt{3}}\right)\frac{1}{\sqrt{3}+5}\)

\(=\left(\frac{2\left(\sqrt{3}+1\right)}{2}+\frac{3\left(\sqrt{3}+2\right)}{-1}+\frac{15\left(3+\sqrt{3}\right)}{6}\right)\frac{\sqrt{3}-5}{-2}\)

\(=\left(\sqrt{3}+1-3\sqrt{3}-6+\frac{45+15\sqrt{3}}{6}\right)\frac{\sqrt{3}-5}{-2}\)

\(=\left(-2\sqrt{3}-5+\frac{45+15\sqrt{3}}{6}\right)\frac{\sqrt{3}-5}{-2}\)

\(=\left(\frac{-12\sqrt{3}-30+45+15\sqrt{3}}{6}\right)\frac{\sqrt{3}-5}{-2}\)

\(=\left(\frac{3\sqrt{3}+15}{6}\right)\frac{5-\sqrt{3}}{2}=\frac{\left(\sqrt{3}+5\right)\left(5-\sqrt{3}\right)}{12}=\frac{22}{12}=\frac{11}{6}\)

a . ta có : \(1\le1+\sqrt{2-x}\Rightarrow GTNN=1\)

\(-2\le\sqrt{x-3}-2\Rightarrow GTNN=-2\)

b. \(0\le\sqrt{4-x^2}\le2\)

\(\sqrt{2x^2-x+3}=\sqrt{2\left(x^2-\frac{x}{2}+\frac{1}{16}\right)+\frac{23}{8}}=\sqrt{2\left(x-\frac{1}{4}\right)^2+\frac{23}{8}}\ge\frac{\sqrt{46}}{4}\)

vậy \(GTNN=\frac{\sqrt{46}}{4}\)

ta có : \(0\le-x^2+2x+5=-\left(x-1\right)^2+6\le6\)

\(\Rightarrow1-\sqrt{6}\le1-\sqrt{-x^2+2x+5}\le1\)Vậy \(\hept{\begin{cases}GTNN=1-\sqrt{6}\\GTLN=1\end{cases}}\)

a, Xét tam giác ABC vuông tại A, đường cao AH

cotC = 7/11 => \(\frac{AB}{AC}=\frac{7}{11}\Rightarrow AB=\frac{7}{11}.AC=\frac{7}{11}.28=\frac{196}{11}\)cm 

Theo định lí Pytago cho tam giác ABC vuông tại A

\(BC=\sqrt{AB^2+AC^2}=\sqrt{\left(\frac{196}{11}\right)^2+28^2}=33,188...\)cm 

b, tanC = 5/7 => \(\frac{AC}{AB}=\frac{5}{7}\Rightarrow AB=\frac{7}{5}AC=\frac{7}{5}.28=\frac{196}{5}\)cm 

Theo định lí Pytago tam giác ABC vuông tại A

\(BC=\sqrt{AB^2+AC^2}=\sqrt{\left(\frac{196}{5}\right)^2+28^2}=\frac{28\sqrt{74}}{5}\)cm 

c, cosC = 4/5 => \(\frac{AC}{BC}=\frac{4}{5}\Rightarrow BC=\frac{5}{4}AC=\frac{5}{4}.28=35\)cm 

Theo định lí Pytago tam giác ABC vuông tại A

\(AB=\sqrt{BC^2-AC^2}=21\)cm 

d, sinC = 3/5 => \(\frac{AB}{BC}=\frac{3}{5}\Rightarrow\frac{AB}{3}=\frac{BC}{5}\Rightarrow\frac{BC^2}{25}=\frac{AB^2}{9}\)

Theo tính chất dãy tỉ số bằng nhau 

\(\frac{BC^2}{25}=\frac{AB^2}{9}=\frac{BC^2-AB^2}{25-9}=\frac{AC^2}{16}=49\)

\(\Rightarrow BC=35cm;AB=21cm\)

hình đơn giản bạn tự vẽ:)

Áp dụng định lý Pytagoras ta có : BC² = AB² + AC² = 3² + 4² = 25 => BC = 5cm

Ta có : \(\sin B=\frac{AC}{BC}=\frac{4}{5};\cos B=\frac{AB}{BC}=\frac{3}{5};\tan B=\frac{AC}{AB}=\frac{4}{3};\cot B=\frac{AB}{AC}=\frac{3}{4}\)

=> \(\sin C=\cos B=\frac{3}{5};\cos C=\sin B=\frac{4}{5};\tan C=\cot B=\frac{3}{4};\cot C=\tan B=\frac{4}{3}\)

Với \(a>0;a\ne1\)

\(P=\left(\frac{1-a\sqrt{a}}{a-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1+a\sqrt{a}}{a+\sqrt{a}}-\sqrt{a}\right)\)

\(=\left(\frac{-\left(\sqrt{a}-1\right)\left(1+\sqrt{a}+a\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}+\sqrt{a}\right)\left(\frac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}-\sqrt{a}\right)\)

\(=\left(\frac{-a-\sqrt{a}-1+a}{\sqrt{a}}\right)\left(\frac{a-\sqrt{a}+1-a}{\sqrt{a}}\right)=\left(\frac{-\sqrt{a}-1}{\sqrt{a}}\right)\left(\frac{1-\sqrt{a}}{\sqrt{a}}\right)\)

\(=-\frac{1-a}{a}=\frac{a-1}{a}\)