mình nghĩ đề như này mới đúng chứ :))

\(\sqrt{2x-5}=\sqrt{5}\)

<=> 2x - 5 = 5

<=> 2x = 5 + 5

<=> 2x = 10

<=> x = 5

\(\sqrt{2x}-5=\sqrt{5}\left(dk:x\ge0\right)\)

\(< =>\sqrt{2x}=\sqrt{5}\left(1+\sqrt{5}\right)\)

\(< =>\sqrt{x}=\frac{\sqrt{5}\left(1+\sqrt{5}\right)}{\sqrt{2}}\)

\(< =>x=\frac{5\left(1+\sqrt{5}\right)^2}{2}\)

Từ 2a + b + c = 0 <=> a + a + b + c = 0 <=> a + c = -(a + b)

Ta có: VT = 2a³ + b³ + c³ = (a³  + b³) + (a³ + c³)

= (a + b)(a² - ab + b²) + (a + c)(a² - ac + c²)

= (a + b)(a² + 2ab + b²) - 3ab(a + b) + (a + c)(a² + 2ac + c²) - 3ac(a + c)

= (a + b)³ - 3ab(a + b) + (a + c)³ - 3ac(a + c)

= (a + b)³ - (a + b)³ - 3ab(a + b) + 3ac(a + b)

= -3a(a + b)(b - c) = 3a(a + b)(c - b) = VP

=> VT = VP => đpcm

Bất đẳng thức cần chứng minh tương đương: \(\frac{1}{a^4\left(b+1\right)\left(c+1\right)}+\frac{1}{b^4\left(c+1\right)\left(a+1\right)}+\frac{1}{c^4\left(a+1\right)\left(b+1\right)}\ge\frac{3}{4}\)

Đặt \(a=\frac{1}{x};b=\frac{1}{y};c=\frac{1}{z}\)thì \(\hept{\begin{cases}x,y,z>0\\xyz=1\end{cases}}\)và ta đưa BĐT cần chứng minh về dạng \(\frac{x^3}{\left(y+1\right)\left(z+1\right)}+\frac{y^3}{\left(z+1\right)\left(x+1\right)}+\frac{z^3}{\left(x+1\right)\left(y+1\right)}\ge\frac{3}{4}\)

Áp dụng BĐT AM - GM, ta được:\(\frac{x^3}{\left(y+1\right)\left(z+1\right)}+\frac{y+1}{8}+\frac{z+1}{8}\ge\frac{3}{4}x\)

Tương tự: \(\frac{y^3}{\left(z+1\right)\left(x+1\right)}+\frac{z+1}{8}+\frac{x+1}{8}\ge\frac{3}{4}y\); \(\frac{z^3}{\left(x+1\right)\left(y+1\right)}+\frac{x+1}{8}+\frac{y+1}{8}\ge\frac{3}{4}z\)

Cộng theo vế của 3 BĐT trên, ta được: \(\frac{x^3}{\left(y+1\right)\left(z+1\right)}+\frac{y^3}{\left(z+1\right)\left(x+1\right)}+\frac{z^3}{\left(x+1\right)\left(y+1\right)}+\)\(\frac{x+y+z+3}{4}\ge\frac{3}{4}\left(x+y+z\right)\)

\(\Rightarrow\frac{x^3}{\left(y+1\right)\left(z+1\right)}+\frac{y^3}{\left(z+1\right)\left(x+1\right)}+\frac{z^3}{\left(x+1\right)\left(y+1\right)}\)\(\ge\frac{1}{2}\left(x+y+z\right)-\frac{3}{4}\ge\frac{1}{2}.3\sqrt[3]{xyz}-\frac{3}{4}=\frac{3}{2}-\frac{3}{4}=\frac{3}{4}\)

Vậy bất đẳng thức được chứng minh

Đẳng thức xảy ra khi x = y = z = 1 hay a = b = c = 1

Đkxđ: x≥0, x khác 49

A= \(\frac{2\sqrt{x}}{\sqrt{x}-7}+\frac{x+21\sqrt{x}}{x-49}\)

A=\(\frac{2\sqrt{x}\left(\sqrt{x}+7\right)+x+21\sqrt{x}}{x-49}\)

=\(\frac{3x+35\sqrt{x}}{x-49}\)

B=\(\frac{\sqrt{x}}{x-5}\)

P=A/B=\(\frac{\sqrt{x}\left(3\sqrt{x}+35\right)\left(x-5\right)}{\sqrt{x}\left(x-49\right)}\)

=\(\frac{\left(3\sqrt{x}+35\right)\left(x-5\right)}{\left(x-49\right)}\)

P=1/3

<=>\(\frac{\left(3\sqrt{x}+35\right)\left(x-5\right)}{\left(x-49\right)}=\frac{1}{3}\)

<=>...

100 dấu căn nha

\(\sqrt{4+\sqrt{4+\sqrt{4+\sqrt{4+...}}}}< \sqrt{6+\sqrt{6+\sqrt{6+...\sqrt{6+\sqrt{9}}}}}\)(100 dấu căn)

=> \(VT< \sqrt{6+\sqrt{6+\sqrt{6+...\sqrt{6+3}}}}=\sqrt{6+\sqrt{6+\sqrt{6+..\sqrt{6+\sqrt{9}}}}}\)(99 dấu căn)

=> \(VT< \sqrt{6+3}=3\)

Trả lời:

\(\sqrt[4]{49+20\sqrt{6}}+\sqrt[4]{49-20\sqrt{6}}=2\sqrt{3}\)

Ta có:\(VT=\sqrt[4]{49+20\sqrt{6}}+\sqrt[4]{49-20\sqrt{6}}\)

                \(=\sqrt[4]{25+20\sqrt{6}+24}+\sqrt[4]{25-20\sqrt{6}+24}\)

                \(=\sqrt[4]{\left(5+2\sqrt{6}\right)^2}+\sqrt[4]{\left(5-2\sqrt{6}\right)^2}\)

                \(=\sqrt{5+2\sqrt{6}}+\sqrt{5-2\sqrt{6}}\)

                \(=\sqrt{3+2\sqrt{6}+2}+\sqrt{3-2\sqrt{6}+2}\)

                \(=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)

                \(=\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}\)

                \(=2\sqrt{3}=VP\) 

Vậy \(\sqrt[4]{49+20\sqrt{6}}+\sqrt[4]{49-20\sqrt{6}}=2\sqrt{3}\)

rộp rộp

\(4x^2+\frac{2x}{\sqrt{x^2+1}+x}-3=0\)

\(\Leftrightarrow4x^2+\frac{2x\left(\sqrt{x^2+1}-x\right)}{\left(\sqrt{x^2+1}+x\right)\left(\sqrt{x^2+1}-x\right)}-3=0\)

\(\Leftrightarrow4x^2+\frac{2x\sqrt{x^2+1}-2x^2}{x^2+1-x^2}-3=0\)

\(\Leftrightarrow2x^2+2x\sqrt{x^2+1}-3=0\)

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

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

Đến đây tự làm , có ý hết rồi 

a) 

Đa thức bậc nhất không phân tích được nhân tử :v

b)

Đặt \(\sqrt{x}=a;\sqrt{y}=b\) Khi đó:

\(x\sqrt{x}+y\sqrt{y}=a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\)

c)

Tương tự câu b) thì ta sẽ có:

\(x\sqrt{x}-27=a^3-27=\left(a-3\right)\left(a^2+3a+9\right)\)

Bài làm:

a) \(9x-5=\left(3\sqrt{x}\right)^2-\sqrt{5}==\left(3\sqrt{x}-\sqrt{5}\right)\left(3\sqrt{x}+\sqrt{5}\right)\)

b) \(x\sqrt{x}+y\sqrt{y}=\left(\sqrt{x}\right)^3+\left(\sqrt{y}\right)^3=\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)\)

c) \(x\sqrt{x}-27=\left(\sqrt{x}\right)^3-3^3=\left(\sqrt{x}-3\right)\left(x+3\sqrt{x}+9\right)\)

Trả lời:

\(\sqrt{x^2-4x+4}-x=-4\)

\(\Leftrightarrow\sqrt{x^2-4x+4}=x-4\)\(\left(ĐK:x\ge4\right)\)

\(\Leftrightarrow x^2-4x+4=\left(x-4\right)^2\)

\(\Leftrightarrow x^2-4x+4=x^2-8x+16\)

\(\Leftrightarrow4x=12\)

\(\Leftrightarrow x=3\left(L\right)\)

Vậy phương trình vô nghiệm 

tam giác BMH đồng dạng với tam giác MCI => \(\frac{BM}{MC}=\frac{MH}{CI}=\frac{BH}{MI}\left(1\right)\)

từ (1) => MB.MC=\(\frac{MH}{CI}\).MC²=\(\frac{MH}{CI}\left(MI^2+IC^2\right)\)=MH.IC+\(\frac{MI}{IC}\cdot MI^2\)

hay MB.MC=IA.IC+\(\frac{BH}{MI}\cdot MI^2\)\(=IA\cdot IA+HB\cdot MI=IA\cdot IC+HB\cdot HA\)