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ko nghĩ mà đòi có kết quả thì bốc cứt

af thế à

\(x^4+x^3+x+1=x^3\left(x+1\right)+\left(x+1\right)=\left(x+1\right)\left(x^3+1\right)=\left(x+1\right)^2\left(x^2-x+1\right)\)

\(x^4-x^3+2x^2-x+1=\left(x^4-x^3+x^2\right)+\left(x^2-x+1\right)=\left(x^2-x+1\right)\left(x^2+1\right)\)

Ta có: \(\left(x+1\right)^2\ge0;\forall x\)

\(x^2+1>1\); \(\forall x\)

\(x^2-x+1=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0,\forall x\)

Vậy \(\frac{x^4+x^3+x+1}{x^4-x^3+2x^2-x+1}=\frac{\left(x+1\right)^2}{x^2+1}\ge0;\forall x\)

Ta có: \(x^2+y^2=\left(x+y\right)^2-2xy=9-2=7\)

\(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=3^3-3.3=18\)

=> \(x^5+y^5=\left(x^2+y^2\right)\left(x^3+y^3\right)-x^2y^2\left(x+y\right)\)

\(=7.18-1.3=123\)

\(A=\left(\frac{2X-1}{x^2-4}+\frac{x+2}{x^2-x-2}\right):\frac{x-2}{x^2+3x+2}ĐK:x\ne\left\{2,-2,-1\right\}\)

a)  \(A=\left[\frac{\left(2x-1\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x+2}{\left(x+1\right)\left(x-2\right)}\right]:\frac{x-2}{\left(x+2\right)\left(x+1\right)}\)

\(A=\left[\frac{\left(2x-1\right)\left(x+1\right)}{\left(x-2\right)\left(x+2\right)\left(x+1\right)}\frac{\left(x+2\right)\left(x+2\right)}{\left(x+1\right)\left(x-2\right)\left(x+2\right)}\right].\frac{\left(x+2\right)\left(x+1\right)}{x-2}\)

\(A=\frac{2x^2+x-1+x^2+4x.4}{\left(x-2\right)\left(x+2\right)\left(x+1\right)}.\frac{\left(x+2\right)\left(x+1\right)}{\left(x-2\right)}\)

\(A=\frac{3x^2+5x+3}{\left(x-2\right)\left(x+2\right)\left(x+1\right)}.\frac{\left(x+2\right)\left(x+1\right)}{\left(x-2\right)}\)

\(A=\frac{3x^2+5x+3}{\left(x-2\right)^2}\)

Ta có :\(3x^2+5x+3\)

\(=3\left(x^2+\frac{5}{3}x+1\right)\)

\(=3\left[x^2+2.\frac{5}{6}x+\frac{25}{36}+\frac{9}{36}\right]\)

\(=3\left[\left(x+\frac{5}{6}\right)^2+\frac{9}{36}\right]>0\)

Mà \(\left(x-2\right)^2>0\)

\(\Rightarrow A>0\left(dpcm\right)\)

\(b,A=11\Leftrightarrow\frac{3x^2+5x+3}{\left(x-2\right)^2}=11\)

\(\Rightarrow3x^2+5x+3=11.\left(x-2\right)^2\)

\(\Rightarrow3x^2+5x+3=11.\left(x^2-4x+4\right)\)

\(\Rightarrow8x^2-49x+41=0\)

\(\Rightarrow8x^2-8x-41x+41=0\)

\(\Rightarrow8x\left(x-1\right)-41\left(x-1\right)=0\)

\(\Rightarrow\left(8x-41\right)\left(x-1\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}8x-41=0\\x-1=0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{41}{8}\\x=1\end{cases}}}\)(Thỏa mãn)

Đặt: x - y = a ; 3x + y - z = b ; -4x + z = c 

Ta có: a + b +  c  = x - y + 3x + y - z - 4x + z = 0 

Khi đó: \(\left(x-y\right)^3+\left(3x+y-z\right)^3+\left(-4x+z\right)^3\)

= \(a^3+b^3+c^3\)

= \(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc+ac\right)+3abc\)

= \(0.\left(a^2+b^2+c^2-ab-bc+ac\right)+3abc\)

= \(3abc\)

= \(3\left(x-y\right)\left(3x+y-z\right)\left(-4x+z\right)\)

cảm ơn ạ 

ĐKXĐ : \(x\ne\pm6\)

\(\frac{36}{x+6}+\frac{36}{x-6}=\frac{9}{2}\)

\(\frac{72\left(x-6\right)}{\left(x+6\right)\left(x-6\right)2}+\frac{72\left(x+6\right)}{\left(x-6\right)\left(x+6\right)2}=\frac{9\left(x+6\right)\left(x-6\right)}{2\left(x+6\right)\left(x-6\right)}\)

\(72\left(x-6\right)+72\left(x+6\right)=9\left(x+6\right)\left(x-6\right)\)

\(72x-432+72x+432=9x^2-324\)

\(144x=9x^2-324\)

\(144x-9x^2+324=0\)

\(-9x^2+144x+324=0\)

\(\Delta=144^2-4.\left(-9\right).324=32400>0\)

Nên phương trình có 2 nghiệm phân biệt 

\(x_1=\frac{-144-\sqrt{32400}}{2.\left(-9\right)}=\frac{-144-180}{-18}=18\)

\(x_2=\frac{-144+\sqrt{32400}}{2.\left(-9\right)}=\frac{-144+180}{-18}=-2\)

Đk : x khác 6 và -6

\(\frac{36}{x+6}+\frac{36}{x-6}=\frac{9}{2}\)

\(< =>\frac{36\left(x-6\right)+36\left(x+6\right)}{\left(x+6\right)\left(x-6\right)}=\frac{9}{2}\)

\(< =>\frac{36x-216+36x+216}{x^2-6x+6x-36}=\frac{9}{2}\)

\(< =>\frac{72x}{x^2-6^2}=\frac{9}{2}\)

\(< =>144x=9x^2-324\)

\(< =>9x^2-144x-324=0\)

Ta có : \(\Delta=\left(-144\right)^2-4.9.\left(-324\right)=32400\)

\(< =>\sqrt{\Delta}=180\)

Vì delta > 0 nên pt có 2 nghiệm phân biệt 

\(x_1=\frac{144+180}{18}=18\)

\(x_2=\frac{144-180}{18}=-2\)

Vậy ...

\(A\ge\frac{\left(x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2}{3}\ge\frac{\left(1+\frac{9}{x+y+z}\right)^2}{3}=\frac{10^2}{3}=\frac{100}{3}\)

ĐTXR ⇔ x = y = z = (x+y+z)/3  = 1/3

\(A^2=\left(x+y\right)^2\le2\left(x^2+y^2\right)=2\Rightarrow-\sqrt{2}\le A\le\sqrt{2}\)

Amin = \(-\sqrt{2}\Leftrightarrow x=y=-\frac{1}{\sqrt{2}}\)

Amax \(=\sqrt{2}\) khi \(x=y=\frac{1}{\sqrt{2}}\)

\(5-2x\left(3-4x^2\right)=5x-8x^2\)

\(5-6x+8x^3=5x-8x^2\)

\(5-6x+8x^3-5x+8x^2=0\)

\(5-11x+8x^3+8x^2=0\)

Bn phân tích nốt.

\(5-2x\left(3-4x^2\right)=5x-8x^2\)

\(< =>5-6x+8x^3=5x-8x^2\)

\(< =>5-6x-5x+8x^3+8x^2=0\)

\(< =>8x^2\left(x+1\right)-11x+5=0\)

Ta có : \(\Delta=\left(-11\right)^2-4.\left(8x+1\right).5=121-160x-20=101-160x\)

đến đây chịu r :(( phải giải pt bậc 3 chăng ?