m = b³+b³

= (a+b ) (a2+b2+ ab) 

mà a+b bằng 1 nên 

m=a2+b2 - ab 

m= (a^2 + b^2 + 2ab ) - 3ab  

3ab = _ < 3 (a+b ) ²/4

=> m _>- 3 (a+b ) ²/4

=1- 3/4  = 3/4 

 chả cần j cả lm bff của nhau thui :3

đù bạn tui dạo nay hok giỏi ghê

\(\sqrt{12}=\sqrt{x^2+12x+13}\)

\(\Leftrightarrow12=x^2+12x+13\)

\(\Leftrightarrow x^2+12x+36-35=0\)

\(\Leftrightarrow\left(x+6\right)^2-35=0\)

\(\Leftrightarrow\left(x+6+\sqrt{35}\right)\left(x+6-\sqrt{35}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=-6-\sqrt{35}\\x=\sqrt{35}-6\end{cases}}\)

Lam thu :3

\(Tk+1=Ck_6.\left(2x\right)^{6-k}.\left(-\frac{1}{x^2}\right)\)

\(=Ck_6.2^{6-k}.x^{6-k}.\frac{\left(-1\right)^k}{x^{2k}}\)

\(-Ck_6.2^{6-k}.x^{6-k-2k}.\left(-1\right)^k\)

SH o chua x \(\Leftrightarrow x^{6-3k}=x^0\)

\(\Leftrightarrow6-3k=0\)

\(\Leftrightarrow k=2\)

\(\Rightarrow SH\)can tim la: \(C^{2_6}.2^4.x^0.\left(-1\right)^2\)

Ta có: \(x^3+y^3\ge xy\left(x+y\right)\Rightarrow1+x^3+y^3\ge xyz+xy\left(x+y\right)\)

\(=xy\left(x+y+z\right)\ge3xy\sqrt[3]{xyz}=3xy\)(vì xyz = 1)

\(\Rightarrow\frac{\sqrt{1+x^3+y^3}}{xy}=\frac{\sqrt{3xy}}{xy}=\sqrt{\frac{3}{xy}}\)

Tương tự ta có: \(\frac{\sqrt{1+y^3+z^3}}{yz}=\sqrt{\frac{3}{yz}}\);\(\frac{\sqrt{1+z^3+x^3}}{zx}=\sqrt{\frac{3}{zx}}\)

Cộng vế với vế, ta được:

\(BĐT=\sqrt{3}\left(\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{zx}}\right)\)

\(\ge3\sqrt{3}\sqrt[3]{\frac{1}{\sqrt{x^2y^2z^2}}}=3\sqrt{3}\)

(Dấu "="\(\Leftrightarrow x=y=z=1\))

\(VT-VP=\Sigma_{cyc}\frac{\frac{1}{2}\left(x+y+1\right)\left(x-y\right)^2}{xy\left(\sqrt{x^3+y^3+1}+\sqrt{3xy}\right)}+\Sigma_{cyc}\frac{\left(x-1\right)^2}{xy\left(\sqrt{x^3+y^3+1}+\sqrt{3xy}\right)}\)

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

\(=\frac{2x-\sqrt{x}-3-x+2\sqrt{x}+8-2+3\sqrt{x}}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}\)

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

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

\(=\frac{\sqrt{x}+3}{\sqrt{x}-4}\)

b) Để \(A\in Z\)

\(\Leftrightarrow\frac{\sqrt{x}+3}{\sqrt{x}-4}=\frac{\sqrt{x}-4}{\sqrt{x}-4}+\frac{7}{\sqrt{x}-4}\in Z\)

=>\(\sqrt{x}-4\inƯ\left(7\right)\)

........

ĐK: \(x\ge1\)

pt <=> \(\sqrt{x-1}+\sqrt{4\left(x-1\right)}-2\sqrt{x-1}=3\)

<=> \(\sqrt{x-1}+2\sqrt{x-1}-2\sqrt{x-1}=3\)

<=> \(\sqrt{x-1}=3\)

<=> x - 1 = 9

<=> x = 10 ( thỏa mãn)

Kết luận: Vậy x = 10.

Đề sai nhé bạn :

Chẳng hạn : \(0+1+2=3\)

Nhưng \(0^2+1^2+2^2=5>3\)nhé 

Đặt \(b=xa;c=ya\Rightarrow a^2+2x^2a^2\le3y^2a^2\Leftrightarrow1+2x^2\le3y^2\)

Ta cần chứng minh:\(\frac{1}{a}+\frac{2}{xa}\ge\frac{3}{ya}\Leftrightarrow1+\frac{2}{x}\ge\frac{3}{y}\)

Vậy ta viết được bài toán thành dạng đơn giản hơn: 

Cho x, y > 0 thỏa mãn \(1+2x^2\le3y^2\). Chứng minh:\(1+\frac{2}{x}\ge\frac{3}{y}\)

Tối về em suy nghĩ tiếp ạ!

Ta co:

\(3c^2\ge a^2+b^2+b^2\ge\frac{\left(a+2b\right)^2}{3}\Rightarrow a+2b\le3c\)

\(\Rightarrow VT=\frac{1}{a}+\frac{1}{b}+\frac{1}{b}\ge\frac{9}{a+2b}\ge\frac{9}{3c}=\frac{3}{c}\)

Dau '=' xay ra khi \(a=b=c=1\)