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

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 ?

gfvfvfvfvfvfvfv555

Vào thống kê hỏi đáp là thấy hình :)

a, 

\(\frac{MF}{MB}=\frac{AF}{BC}=\frac{AD-DF}{BC}\)

\(=1-\frac{ED}{EC}=\frac{EC-ED}{EC}=\frac{DC}{EC}=\frac{AB}{EC}=\frac{MB}{ME}\)

\(\Rightarrow MB^2=MF.ME\)

b,

\(\frac{1}{BE}+\frac{1}{BF}=\frac{1}{BM}\Leftarrow BM\left(BE+BF\right)=BE.BF\Leftarrow BM.BF=BE.\left(BF-BM\right)=BE.BF\Leftarrow\frac{BE}{BM}\)
\(=\frac{BF}{MF}\Leftarrow\frac{ME}{MB}=\frac{MB}{MF}\)

Nguồn : gg

\(\left(x^2+x+1\right)\left(x^2+x+2\right)=12\)

\(x^4+x^3+2x^2+x^3+x^2+2x+x^2+x+2=12\)

\(x^4+2x^3+4x^2+3x+2=12\)

\(x^4+2x^3+4x^2+3x+2-12=0\)

\(x^4+2x^3+4x^2+3x-10=0\)

\(\left(x^2+x+5\right)\left(x+2\right)\left(x-1\right)=0\)

TH1 : \(x^2+x+5=0\)

\(\Delta=1^2-4.1.5=1-20=-19< 0\)

Nên phương trình vô nghiệm.

TH2 : \(x+2=0\Leftrightarrow x=-2\)  

TH3 : \(x-1=0\Leftrightarrow x=1\)

\(\left(x^2+x+1\right)\left(x^2+x+2\right)=12\)

Đặt \(x^2+x+1=t\)

\(\Rightarrow t\left(t+1\right)=12\)\(\Leftrightarrow t^2+t=12\)

\(\Leftrightarrow t^2+t-12=0\)\(\Leftrightarrow\left(t^2-3t\right)+\left(4t-12\right)=0\)

\(\Leftrightarrow t\left(t-3\right)+4\left(t-3\right)=0\)\(\Leftrightarrow\left(t-3\right)\left(t+4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}t-3=0\\t+4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}t=3\\t=-4\end{cases}}\)

Ta thấy: \(x^2+x+1=x^2+2.\frac{1}{2}x+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)

\(\Rightarrow t>0\)\(\Rightarrow t=3\)thoả mãn

\(\Rightarrow x^2+x+1=3\)\(\Leftrightarrow x^2+x+1-3=0\)\(\Leftrightarrow x^2+x-2=0\)

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

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-2\end{cases}}\)

Vậy tập nghiệm của phương trình là \(S=\left\{-2;1\right\}\)

Gọi quãng đường AB là x ( x>0 )

Thời gian đi \(\frac{1}{3}\)quãng đường đầu là \(\frac{1}{3}x:40=\frac{x}{120}h\)

Thời gian đi \(\frac{2}{3}\)quãng đường sau là \(\frac{2}{3}x:50=\frac{2x}{150}=\frac{x}{75}h\)

Vì tổng thời gian hết 7h nên ta có phương trình

\(\frac{x}{120}+\frac{x}{75}=7\)

Còn đâu bạn tự tính phương trình nốt nha