\(x^2+2x\sqrt{x+\frac{1}{x}}=8x-1\)(đk;x>0)

\(\Leftrightarrow x^2+2\sqrt{x}\cdot\sqrt{x^2+1}=8x-1\)

\(\Leftrightarrow\left(x^2+1\right)+2\sqrt{x}\cdot\sqrt{x^2+1}+x=9x\)

\(\Leftrightarrow\left(\sqrt{x^2+1}+\sqrt{x}\right)^2-9x=0\)

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

\(\Leftrightarrow\left(\sqrt{x^2+1}+4\sqrt{x}\right)\left(\sqrt{x^2+1}-2\sqrt{x}\right)=0\)

\(\Leftrightarrow\sqrt{x^2+1}-2\sqrt{x}=0\)(vì \(\sqrt{x^2+1}+4\sqrt{x}>0\))

\(\Leftrightarrow x^2-4x+1=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=2-\sqrt{3}\\x=2+\sqrt{3}\end{cases}}\)(thõa mãn điều kiện)

\(\sqrt{x-2009}-\sqrt{y-2008}-\sqrt{z-2}=\frac{1}{2}\left(x+y+z\right)\)(đk:x>2009,y>2008,z>2)

\(\Leftrightarrow\left(\sqrt{x-2009}-1\right)^2+\left(\sqrt{x-2008}+1\right)^2+\left(\sqrt{z-2}+1\right)^2+4014=0\)(không thõa mãn)

Lý do có kết quả trên là vì chuyển 1\2 qua vế trái và tách theo hằng đẳng thức

Bài tiếp theo cũng làm tương tự

Đặt x=a-2,ta có : \(P=\frac{\sqrt{x}-2}{3}.\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{9-x}\right):\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}-\frac{1}{\sqrt{x}}\right)\)

\(=\frac{\sqrt{x}-2}{3}.\left(\frac{\sqrt{x}\left(3-\sqrt{x}\right)+x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right):\left(\frac{3\sqrt{x}+1-\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\)

\(=\frac{\sqrt{x}-2}{3}.\left(\frac{3\left(\sqrt{x}+3\right)}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right):\left(\frac{2\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\)

\(=\frac{\sqrt{x}-2}{3}.\frac{3}{3-\sqrt{x}}.\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\left(\sqrt{x}+2\right)}\)

\(=\frac{-\sqrt{x}\left(\sqrt{x}-2\right)}{2\left(\sqrt{x}+2\right)}\)

Hình dễ vẽ , bạn tự vẽ nha.

a) \(\Delta AKH\)vuông tại A có \(AK^2=KE.KH\)hay \(6^2=KE.10\Rightarrow KE=3,6\)

Vậy KE=3,6

a) \(P=\frac{3\left(x+\sqrt{x}-3\right)}{x+\sqrt{x}-2}+\frac{\sqrt{x}+3}{\sqrt{x}+2}-\frac{\sqrt{x}-2}{\sqrt{x}-1}\) \(\left(x\ge0;x\ne1\right)\)

\(P=\frac{3x+3\sqrt{x}-9}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}+\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(P=\frac{3x+3\sqrt{x}-9+x+2\sqrt{x}-3-x+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(P=\frac{3x+5\sqrt{x}-8}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(P=\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+8\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(P=\frac{3\sqrt{x}+8}{\sqrt{x}+2}\)

b) \(P=\frac{7}{2}\)

\(\Leftrightarrow\frac{3\sqrt{x}+8}{\sqrt{x}+2}=\frac{7}{2}\)

\(\Rightarrow6\sqrt{x}+16=7\sqrt{x}+14\)

\(\Leftrightarrow\sqrt{x}=2\Rightarrow x=4\)

ta có \(A=\frac{3+\sqrt{5}}{4+\sqrt{2\left(3+\sqrt{5}\right)}}=\frac{3+\sqrt{5}}{4+\sqrt{6+2\sqrt{5}}}=\frac{3+\sqrt{5}}{4+\sqrt{\left(\sqrt{5}+1\right)^2}}=\frac{\left(3+\sqrt{5}\right)}{5+\sqrt{5}}\)\(=\frac{\left(5-\sqrt{5}\right)\left(3+\sqrt{5}\right)}{20}=\frac{5+\sqrt{5}}{10}\)

tương tự \(B=\frac{3-\sqrt{5}}{4-\sqrt{2\left(3-\sqrt{5}\right)}}=\frac{5-\sqrt{5}}{10}\)

\(\Rightarrow A-B=\frac{\sqrt{5}}{5},A+B=1;AB=\frac{1}{5}\)

vậy \(A^3-B^3=\left(A-B\right)\left(A^2+AB+B^2\right)=\left(A+B\right)\left[\left(A+B\right)^2-AB\right]=\frac{\sqrt{5}}{5}\left(1-\frac{1}{5}\right)\cdot\frac{4}{5}=\frac{4\sqrt{5}}{25}\)

\(=\frac{\sqrt{5}\left(\sqrt{2}-\sqrt{3}\right)}{\sqrt{4}\left(\sqrt{2}-\sqrt{3}\right)}\)

\(=\frac{\sqrt{5}}{2}\)

Do \(OM\perp PQ\Rightarrow\) M la diem giua cung PQ

=> EM la phan giac goc PEQ

ma EM vuong goc EN ( MN la duong kinh )

=> EN la phan giac ngoai goc PEQ

khi do ta co \(\frac{NS}{NH}=\frac{MS}{MH}\Rightarrow\frac{NS}{MS}=\frac{NH}{MH}\)

suy ra \(\frac{NS}{NS+MS}=\frac{NH}{NH+MH}=\frac{NH}{MN}\Rightarrow NH=\frac{NS.MN}{NS+MS}=const\) (Do M,N S co dinh )

suy ra N co dinh ma O co dinh nen \(OH=const\left(dpcm\right)\)

đường kính MN nhé các bạn , mình đánh hơi ẩu

Theo định lý côsin ta có \(a^2=b^2+c^2-2bc.cosA\)

Khi  \(a^2=b^2+c^2-bc\)thì \(2cosA=1\Rightarrow cosA=\frac{1}{2}\Rightarrow\widehat{A}=60^o\)

Khi \(a^2=b^2+c^2+bc\) thì \(-2cosA=1\Rightarrow cosA=-\frac{1}{2}\)(Khúc này để chứng minh ∠A = 120o khi và chỉ khi a 2 = b 2 + c 2 + bc. mà nó ra vầy mik chịu á , bn xem lại đề ik nha)

tai sao cosA =1/2 thi goc A lai bang 60o vay bn

a) \(\sqrt{8-\sqrt{60}}\)=\(\sqrt{8-\sqrt{4.15}}\)=\(\sqrt{8-2\sqrt{15}}\)=\(\sqrt{\left(\sqrt{5}\right)^2-2\sqrt{5}\sqrt{3}+\left(\sqrt{3}\right)^2}\)=\(\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)=l\(\sqrt{5}\)\(-\sqrt{3}\)l =\(\sqrt{5}\)\(-\sqrt{3}\)(do \(\sqrt{5}\)\(-\sqrt{3}\)>0)

Các câu còn lại bạn làm tương tự câu a là ra

a) \(\sqrt{27x^2}=\sqrt{3.\left(3x\right)^2}=\left|3x\right|.\sqrt{3}=3x\sqrt{3}\left(x>0\right)\)

b) \(\sqrt{8xy^2}=\left|y\right|.2\sqrt{2x}=-2y\sqrt{2x}\left(x\ge0,y\le0\right)\)

1) \(x\sqrt{13}=\sqrt{13x^2}\left(x\ge0\right)\)

2) \(x\sqrt{-15x}=-\left|x\right|\sqrt{15x}=-\sqrt{15x^3}\left(x< 0\right)\)

3) \(x\sqrt{2}=-\left|x\right|\sqrt{2}=-\sqrt{2x^2}\left(x\le0\right)\)