phan h da thuc tren thanh nhan tu

   \(\left(2x+5\right)^2-6x-15\)

\(=4x^2+20x+25-6x-15\)

\(=4x^2+14x+10\)

\(=2\left(2x^2+7x+10\right)\)

\(\left(2x+5\right)^2-6x-15=\left(2x+5\right)^2-3\left(2x-5\right)=\left(2x-5\right)^2-3\left(2x-5\right)\)

                                                \(=\left(2x-5\right)\left(2x-5-3\right)=\left(2x-5\right)\left(2x-2\right)=2\left(2x-5\right)\left(x-1\right)\)

(2x+5)^2-6x-15

=4x²+20x+25-6x-15

=4x²+14x+10

=(2x+1)²+9        (đề bài có nhầm không)

Em tham khảo link:Câu hỏi của Conan Kudo - Toán lớp 8 - Học toán với OnlineMath

Ta có bổ đề

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)\(\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)

ÁP DỤNG BỔ ĐỀ VÀO P ta có

\(P=\frac{bc}{a^2}+\frac{ca}{b^2}+\frac{ab}{c^2}=abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)\)

\(=abc.\frac{3}{abc}=3\)

Vậy P=3

Từ giả thiết: \(\frac{a}{b}=\frac{b}{c}\Rightarrow ac=b^2\Rightarrow abc=b^3\)

Ta có: \(\frac{a^3-2b^3+c^3}{a+b+c}=\frac{a^3+b^3+c^3-3c^3}{a+b+c}=\frac{a^3+b^3+c^3-3abc}{a+b+c}\)

Xét: \(a^3+b^3+c^3=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)

\(=\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)\)

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

\(\Rightarrow\frac{a^3-2b^3+c^3}{a+b+c}=a^2+b^2+c^2-ab-bc-ac\) là 1 số nguyên (đpcm)

Sai r bạn ơi

Đặt x=a + b - 2c
       y=b+c-2a
       z=c+a-2b
=>x+y+z=(a + b - 2c)+(b+c-2a)+(c+a-2b)
=>x+y+z=0
=>x+y= - z                         (1)
=>(x+y)^3=(-z)^3
=>x^3+y^3+3xy(x+y)=(-z)^3
=>x^3+y^3+z^3 +3xy(-z)=0        {vì x+y=-z [theo (1)]}
=>x^3+y^3+z^3 -3xyz=0
=>x^3+y^3+z^3 =3xyz
Vậy (a + b - 2c)^3 + (b + c - 2a)^3 + (c + a - 2b)^3=3(a + b - 2c) (b + c - 2a)(c + a - 2b)

\(\left(x-1\right)^3+\left(x-3\right)^3+8\left(2-x\right)^3=0\)

\(\left(x-1+x-3\right)\left[\left(x-1\right)^2-\left(x-1\right)\left(x-3\right)+\left(x-3\right)^2\right]+\left[2\left(2-x\right)\right]^3=0\)

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

\(\left(2x-4\right)\left(x^2-4x+7\right)-\left(2x-4\right)^3=0\)

\(\left(2x-4\right)\left[x^2-4x+7-\left(2x-4\right)^2\right]=0\)

\(2\left(x-2\right)\left(x^2-4x+7-4x^2+16x-16\right)=0\)

\(2\left(x-2\right)\left(12x-3x^2-9\right)=0\)

\(6\left(x-2\right)\left(4x-x^2-3\right)=0\)

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

\(6\left(x-2\right)\left[x\left(3-x\right)-\left(3-x\right)\right]=0\)

\(6\left(x-2\right)\left(3-x\right)\left(x-1\right)=0\)

\(\Rightarrow x=\left\{1;2;3\right\}\)

\(\left(x-1\right)^3+\left(x-3\right)^3+8\left(2-x\right)^3=0\)

\(\Rightarrow x^3-2x^2+x-x^2+2x+1+x^3-6x^2+9x-3x^2+18x-27+64-64x+16x^2-32x+32x^2-8x^3=0\)

\(\Rightarrow-6x^3+36x^2-66x+36=0\)

\(\Rightarrow-6\left(x^3-6x^2+11x-6\right)=0\)

\(\Rightarrow\left(x^2-5x+6\right)\left(x-1\right)=0\)

\(\Rightarrow\left(x-3\right)\left(x-2\right)\left(x-1\right)=0\)

=> x - 3 = 0 ; x - 2 = 0 hoặc x - 1 = 0

=> x = 3 ; x = 2 hoặc x = 1

a) 16(4x+5)² - 25(2x+2)²

^{\(=\left[4\left(4x+5\right)\right]^2-\left[5\left(2x+2\right)\right]^2\)}

^{\(=\left[4\left(4x+5\right)+5\left(2x+2\right)\right]\left[4\left(4x+5\right)-5\left(2x+2\right)\right]\)}

^{\(=\left(16x+20+10x+10\right)\left(16x+20-10x-10\right)\)}

^{\(=\left(26x+30\right)\left(6x+10\right)\)}

\(b,\left(x-y+4\right)^2-\left(2x+3y-1\right)^2\)

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

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

\(c,\left(x+1\right)^4-\left(x-1\right)^4\)

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

\(=\left[\left(x+1\right)^2+\left(x-1\right)^2\right]\left[\left(x+1\right)^2-\left(x-1\right)^2\right]\)

\(=\left(x^2+2x+1+x^2-2x+1\right)\left[\left(x+1+x-1\right)\left(x+1-x+1\right)\right]\)

\(=\left(2x^2+2\right)2x.2\)

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

\(=8x\left(x^2+1\right)\)

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

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

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

\(\Rightarrow\left(x-2\right)3x=0\)

\(\Rightarrow\orbr{\begin{cases}3x=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)