\(ĐKXĐ:x\ne\pm1\)

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

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

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

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

\(\Leftrightarrow-3x^4-4x^3-x^2+4x+1=0\)

a) Ta có :

\(\frac{AE}{AB}=\frac{1,5}{6}=\frac{1}{4}\)

\(\frac{AF}{AC}=\frac{2}{8}=\frac{1}{4}\)

\(\Rightarrow\frac{AE}{AB}=\frac{AF}{AC}\)

\(\Rightarrow EF//BC\)(Theo định lí Ta-lét đảo)

b)Áp dụng định lí Pythagoras vào △ABC vuông tại A :

         BC² = AB² + AC²

\(\Rightarrow\)BC² = 6² + 8²

\(\Rightarrow\)BC² = 100

\(\Rightarrow\)BC   = 10 cm

Xét △ABC có : MN // BC

\(\Rightarrow\frac{AE}{AB}=\frac{AF}{AC}=\frac{EF}{BC}\)(Hệ quả định lí Ta-lét)

\(\Rightarrow\frac{EF}{BC}=\frac{1}{4}\)

\(\Rightarrow EF=\frac{1}{4}BC=\frac{1}{4}\cdot10=2,5\left(cm\right)\)

c) Xét △KBC có EF // BC

\(\Rightarrow\frac{KB}{KF}=\frac{KC}{KE}\)(Theo định lí Ta-lét)

\(\Rightarrow KE.KB=KF.KC\)

Đề bị nhầm hay sao ý? Cậu ktra lại nhé

ko nhàm đâu

Phân số đó là \(\frac{9}{20}\)

\(A=\left(3+1\right)\left(3^2+1\right)...\left(3^{64}+1\right)\)

\(\Leftrightarrow\left(3-1\right)A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)...\left(3^{64}+1\right)\)

\(\Leftrightarrow2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(\Leftrightarrow2A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)...\left(3^{64}+1\right)\)

\(\Leftrightarrow2A=3^{128}-1\)

\(\Leftrightarrow A=\frac{3^{128}-1}{2}\)

ko hieu

(2x-7)^2-6(2x-7)(x-3)=0

=> (2x-7)(2x-7-6x+18)=0

=>(2x-7)(11-4x)=0

=>2x-7=0 hoặc 11-4x=0

+) Nếu 2x-7=0 thì 2x=7 nên x=7/2

+) Nếu 11-4x=0 thì 4x=11 nên x=11/4

 Nhớ k cho mình với nha, mình trả lời đầu tiên.