\(1+6x-6x^2-x^3=0\)

\(\Leftrightarrow\left(1-x\right)^3=0\)

\(\Leftrightarrow1-x=0\)

\(\Leftrightarrow x=1\)

Cmr :\(\frac{a^n+b^n}{a^{n-1}+b^{n-1}}\ge\frac{a^{n-1}+b^{n-1}}{a^{n-2}+b^{n-2}}\) với n > 1 và n ∈ N

help !

\(\left(-169\right)+72\left(-169\right)+\left(-169\right).27\)

\(=\left(-169\right).\left(1+72+27\right)\)

\(=\left(-169\right).100\)

\(=-16900\)

\(x^4-4x^3+8x^2-16x+16\)

\(=\left(x^4+8x^2+16\right)-\left(4x^3+16x\right)\)

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

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

a, Điều kiện của x để A được xác định là:

\(x-2\ne0\)

=> \(x\ne2\)

Vậy..........

b, Với \(x\ne2\)

\(A=\dfrac{3x^2-6x}{x-2}=\dfrac{3x\left(x-2\right)}{x-2}=3x\)

Thay x = 4 (tmđk) vào A ta được:

3.4 = 12

Vậy............

a, Vì M là trung điểm cạnh BC => MB = MC

Xét △ABM và △ACM có:

AB = AC (gt)

MB = MC (cmt)

AM chung

=> △ABM = △ACM (c-c-c)

b, Vì △ABM = △ACM (cmt)

=> \(\widehat{AMB}=\widehat{AMC}\) (2 góc tương ứng)

mà \(\widehat{AMB}+\widehat{AMC}=180^o\)

=> \(2\widehat{AMB}=180^o\)

=> \(\widehat{AMB}=90^o\)

=> AM ⊥ BC

c, Xét △ADM và △AEM có:

AD = AE(gt)

\(\widehat{A_1}=\widehat{A_2}\) (do ABM = ACM)

AM chung

=> △ADM = △AEM (c-g-c)

a, \(\dfrac{4x+13}{5x\left(x-7\right)}-\dfrac{x-48}{5x\left(7-x\right)}\)

\(=\dfrac{4x+13}{5x\left(x-7\right)}+\dfrac{x-48}{5x\left(x-7\right)}\)

\(=\dfrac{4x+13+x-48}{5x\left(x-7\right)}\)

\(=\dfrac{5x-35}{5x\left(x-7\right)}\)

\(=\dfrac{5\left(x-7\right)}{5x\left(x-7\right)}=\dfrac{1}{x}\)

b, \(\dfrac{1}{x-5x^2}-\dfrac{25x-15}{25x^2-1}\)

\(=\dfrac{1}{x\left(1-5x\right)}+\dfrac{25x-15}{\left(1-5x\right)\left(1+5x\right)}\)

\(=\dfrac{1+5x}{x\left(x-5x\right)\left(1+5x\right)}+\dfrac{x\left(25x-15\right)}{x\left(1-5x\right)\left(1+5x\right)}\)

\(=\dfrac{1+5x+25x^2-15x}{x\left(1-5x\right)\left(1+5x\right)}\)\(=\dfrac{25x^2-10x+1}{x\left(1-5x\right)\left(1+5x\right)}=\dfrac{\left(5x-1\right)^2}{x.\left(1-5x\right)\left(1+5x\right)}\)

\(=\dfrac{\left(5x-1\right)^2}{-x\left(5x-1\right)\left(1+5x\right)}\) \(=\dfrac{-\left(5x-1\right)}{x\left(1+5x\right)}\)