\(M=\left(3x-4\right)\left(9x^2-12x+16\right)+\left(6x+8\right)^2\)

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

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

\(\Rightarrow M=\left(3x-4\right)^2.3x\)

Ko ghi lại đề

\(=>M=\left(3X-4\right)\left(3x-4\right)^2+4\left(3x-4\right)^2\)

\(=>M=\left(3X-4\right)^2\left(3x-4+4\right)\)

\(=>M=\left(3X-4\right)^2\left(3x\right)\)

~Study well~ :)

ai trả lời nhanh câu này mình sẽ tích cho

\(\left(x-1\right)^3-\left(x+1\right)^3+6\left(x+1\right)\left(x-1\right)\)

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

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

\(=-6x^2-2+6x^2-6\)

\(=-8\)

Vậy giá trị của của biểu thức không phụ thuộc vào biến x.

\(a,\left(x-2\right)^3-x\left(x+1\right)\left(x-1\right)+6x\left(x-3\right)\)

\(=\left(x-2\right)^3-x\left(x^2-1\right)+6x^2-18x\)

\(=x^3-6x^2+12x-8-x^3+x+6x^2-18x\)

\(=-5x\)

Các câu còn lại lm tương tự nhé

Mình giải đc bài 2 rồi các bạn chỉ cần giúp mình bài 1

ms hok hả bn?

\(H_2+O_2\rightarrow H_2O\)

P/s :cs điều kiện nhiệt độ trên dấu mũi tên

á wên cân bằng,hệ số cân = là:2:1:2

\(\frac{5}{a}+\frac{3}{a+4}=\frac{5.\left(a+4\right)+3a}{a.\left(a+4\right)}=\frac{5a+20+3a}{a^2+4a}\)

                          \(=\frac{8a+20}{a^2+4a}\)

\(\frac{4}{c-5}+\frac{2}{2c+3}\) \(=\frac{4\left(2c+3\right)+2\left(c-5\right)}{\left(c-5\right)\left(2c+3\right)}\)

                                        \(=\frac{8c+12+2c-10}{2c^2+3c-10c-15}\)

                                         \(=\frac{10c-2}{2c^2-7c-15}\)

câu còn lại tương tự nha

mk phải đi học rồi

Ta có :  \(\frac{x^2}{y^2}+\frac{y^2}{x^2}+4-\frac{x}{y}-\frac{y}{x}\) \(\left(1\right)\)

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

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

Mà  \(\left(\frac{x}{y}-\frac{1}{2}\right)^2\ge0\)

       \(\left(\frac{y}{x}-\frac{1}{2}\right)^2\ge0\)

\(\Rightarrow\left(1\right)\ge\frac{1}{2}>0\)( luôn đúng ) 

\(\Rightarrow\frac{x^2}{y^2}+\frac{y^2}{x^2}+4>\frac{x}{y}+\frac{y}{x}\left(đpcm\right)\)

1.a) (3x+1)²-4(x-2)²= (3x+1)²-[2(x-2)]²=[(3x+1)-2(x-2)][(3x+1)+2(x-2)]=(x+3)(5x-1)

b) (a²+b²-5)²-4(ab+2)²= (a²+b²-5)²-[2(ab+2)]² = (a²+b²-5-2ab-4)(a²+b²-5+2ab+4)=[(a-b)²-9][(a+b)²-1]

2. 3x²+9x-30=3x²-6x+15x-30=3x(x-2)+15(x-2)=3(x+5)(x-2)

b. x³-5x²-14x=x³+2x²-7x²-14x=x²(x+2)-7x(x+2)=(x²-7x)(x+2)

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

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

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

\(=\left[3x+1-2x+4\right].\left[3x+1+2x-4\right]\)

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

b) \(\left(a^2+b^2-5\right)^2-4\left(ab+2\right)^2\)

\(=\left(a^2+b^2-5\right)^2-\left[2.\left(ab+2\right)\right]^2\)

\(=\left(a^2+b^2-5\right)^2-\left(2ab+4\right)^2\)

\(=\left(a^2+b^2-5-2ab-4\right)\left(a^2+b^2-5+2ab+4\right)\)

\(=\left[\left(a-b\right)^2-9\right].\left[\left(a+b\right)^2-1\right]\)

\(=\left[\left(a-b-3\right)\left(a-b+3\right)\right].\left[\left(a+b-1\right)\left(a+b+1\right)\right]\)

a) \(3x^2+9x-30\)

\(=3\left(x^2+3x-10\right)\)

\(=3\left(x^2-2x+5x-10\right)\)

\(=3.\left[x\left(x-2\right)+5.\left(x-2\right)\right]\)

\(=3.\left[\left(x+5\right)\left(x-2\right)\right]\)

b) \(x^3-5x^2-14x\)

\(=x\left(x^2-5x-14\right)\)

\(=x\left(x^2+2x-7x-14\right)\)

\(=x.\left[x\left(x+2\right)-7.\left(x+2\right)\right]\)

\(=x.\left[\left(x-7\right)\left(x+2\right)\right]\)

a) 4x²-1=(2x)²-1²=(2x-1)(2x+1)

b)25x²-0.09=(5x)²-\(\left(\frac{3}{10}\right)^2\)=\(\left(5x-\frac{3}{10}\right)\left(5x+\frac{3}{10}\right)\)

c)9x²-14=(3x)²-\(\sqrt{14}^2\)=(3x-\(\sqrt{14}\))(3x+\(\sqrt{14}\))

d) (x-y)²-4=(x-y)²-2²=(x-y-2)(x-y+2)

e) 9-(x-y)²=3³-(x-y)²=(3-x+y)(3+x-y)

f)(x²+4)²-16x²=(x²+4)²-(4x)²=(x²+4+4x)(x²+4-4x)

Chúc hok tốt!!!

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

\(b,\)\(25x^2-0,09=\left(5x\right)^2-0,3^2=\left(5x-0,3\right)\left(5x+0,3\right)\)

\(c,\)\(9x^2-14=\left(3x\right)^2-\left(\sqrt{14}\right)^2=\left(3x-\sqrt{14}\right)\left(3x+\sqrt{14}\right)\)

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

\(e,\)\(9-\left(x-y\right)^2=3^2-\left(x-y\right)^2=\left(3-x+y\right)\left(3+x-y\right)\)

\(f,\)\(\left(x^2+4\right)^2-16x^2=\left(x^2+4\right)^2-\left(4x\right)^2=\left(x^2+4-4x\right)\left(x^2+4+4x\right)\)

                                             \(=\left(x^2-2.2x+2^2\right)\left(x^2+2.2x+2^2\right)\)

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