Vì dài quá nên mình chỉ có thể trả lời được mấy câu thôi

Bài 1:

27x³ - 8 : (6x + 9x² +4)

= (3x - 2) (9x² + 6x + 4) : (9x² + 6x + 4)

= 3x - 2

Bài 3:

a, 81x⁴ + 4 = (9x²)² + 36x² + 4 - 36x²

= (9x² + 2)² - (6x)²

= (9x² + 6x + 2)(9x² - 6x + 2)

b, x² + 8x + 15 = x² + 3x + 5x + 15

= x(x + 3) + 5(x + 3)

= (x + 3)(x + 5)

c, x² - x - 12 = x² + 3x - 4x - 12

= x(x + 3) - 4(x + 3)

= (x + 3) (x - 4)

Câu 1:

(27x³ - 8) : (6x + 9x² + 4)

= (3x - 2)(9x² + 6x + 4) : (6x + 9x² + 4)

= 3x - 2

Câu 2:

a) (3x - 5)(2x+ 11) - (2x + 3)(3x + 7)

= 6x² + 33x - 10x - 55 - 6x² - 14x - 9x - 21

= -76

⇒ đccm

b) (2x + 3)(4x² - 6x + 9) - 2(4x³ - 1)

= 8x³ + 27 - 8x³ + 2

= 29

⇒ đccm

Câu 3:

a) 81x⁴ + 4

= (9x²)² + 2²

= (9x² + 2)² - (6x)²

= (9x² - 6x + 2)(9x² + 6x + 2)

b) x² + 8x + 15

= x² + 3x + 5x + 15

= x(x + 3) + 5(x + 3)

= (x + 3)(x + 5)

c) x² - x - 12

= x² - 4x + 3x - 12

= x(x - 4) + 3(x - 4)

= (x - 4)(x + 3)

c/m biểu thức không phụ thuộc vào biến

a: \(=\dfrac{x^2+2x+1+6-x^2-2x+3}{2\left(x-1\right)\left(x+1\right)}\cdot\dfrac{2\left(x-1\right)\left(x+1\right)}{5}\cdot2\)

\(=\dfrac{10}{5}\cdot2=4\)

b: \(=\dfrac{x}{x-3}-\dfrac{x\left(x+3\right)}{2x+3}\cdot\dfrac{x^2+6x+9-x^2}{x\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{x}{x-3}-\dfrac{3}{x-3}=1\)

a) \(\dfrac{x+1}{2}+\dfrac{3x-2}{3}=\dfrac{x-7}{12}\)

\(\Leftrightarrow\dfrac{6\left(x+1\right)+4\left(3x-2\right)}{12}=\dfrac{x-7}{12}\)

\(\Leftrightarrow6\left(x+1\right)+4\left(3x-2\right)=x-7\)

\(\Leftrightarrow6x+6+12x-8=x-7\)

\(\Leftrightarrow6x+12x-x=-7-6+8\)

\(\Leftrightarrow17x=-5\)

\(\Leftrightarrow x=\dfrac{-5}{17}\)

Vậy .........................

b) \(\dfrac{2x}{x-3}-\dfrac{5}{x+3}=\dfrac{x^2+21}{x^2-9}\left(ĐKXĐ:x\ne\pm3\right)\)

\(\Leftrightarrow\dfrac{2x\left(x+3\right)-5\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{x^2+21}{\left(x-3\right)\left(x+3\right)}\)

\(\Rightarrow2x\left(x+3\right)-5\left(x-3\right)=x^2+21\)

\(\Leftrightarrow2x^2+6x-5x+15=x^2+21\)

\(\Leftrightarrow2x^2-x^2+x+15-21=0\)

\(\Leftrightarrow x^2+x-6=0\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(n\right)\\x=-3\left(l\right)\end{matrix}\right.\)

Vậy \(S=\left\{2\right\}\)

d) \(\left(x-4\right)\left(7x-3\right)-x^2+16=0\)

\(\Leftrightarrow\left(x-4\right)\left(7x-3\right)-\left(x^2-16\right)=0\)

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

\(\Leftrightarrow\left(x-4\right)\left(7x-3-x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(6x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\6x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{7}{6}\end{matrix}\right.\)

Vậy .........................

P/s: các câu còn lại tương tự, bn tự giải nha

GIÚP MÌNH VỚI MAI LÀ NỘP BÀI RỒI

câu a) và b) thì sử dụng tính chất nếu tích =0 thì có ít nhất 1 thừa số =0

c)4x^2+4x+1=0

(2x+1)^2=0

2x+1=0

x=-1/2

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

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

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

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

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

Thay \(x=\dfrac{3}{2}\) vào biểu thức A ta được:

\(\left(3-2.\dfrac{3}{2}\right)^2=\left(3-3\right)^2=0^2=0\)

Vậy giá trị của biểu thức A tại \(x=\dfrac{3}{2}\) là 0

b) \(B=\dfrac{x^2y\left(y-x\right)-xy^2\left(x-y\right)}{3y^2-3x^2}\)

\(\Leftrightarrow B=\dfrac{x^2y\left(y-x\right)+xy^2\left(y-x\right)}{3\left(y^2-x^2\right)}\)

\(\Leftrightarrow B=\dfrac{\left(y-x\right)\left(x^2y+xy^2\right)}{3\left(y-x\right)\left(y+x\right)}\)

\(\Leftrightarrow B=\dfrac{xy\left(y-x\right)\left(x+y\right)}{3\left(y-x\right)\left(y+x\right)}\)

\(\Leftrightarrow B=\dfrac{xy\left(y-x\right)\left(y+x\right)}{3\left(y-x\right)\left(y+x\right)}\)

\(\Leftrightarrow B=\dfrac{xy}{3}\)

Thay \(x=-3\) và \(y=\dfrac{1}{2}\) vào biểu thức B ta được:

\(\dfrac{\left(-3\right).\dfrac{1}{2}}{3}=\dfrac{\dfrac{-3}{2}}{3}=\dfrac{\dfrac{-3}{2}}{3}=\dfrac{-1}{2}\)

Vậy giá trị của biểu thức B tại \(x=-3\) và \(y=\dfrac{1}{2}\) là \(\dfrac{-1}{2}\)

c) \(C=\dfrac{x+1}{x-3}-\dfrac{1-x}{x+3}-\dfrac{2x\left(1-x\right)}{9-x^2}\)

\(\Leftrightarrow C=\dfrac{x+1}{x-3}-\dfrac{1-x}{x+3}+\dfrac{2x\left(1-x\right)}{x^2-9}\)

\(\Leftrightarrow C=\dfrac{x+1}{x-3}-\dfrac{1-x}{x+3}+\dfrac{2x\left(1-x\right)}{\left(x-3\right)\left(x+3\right)}\) MTC: \(\left(x-3\right)\left(x+3\right)\)

\(\Leftrightarrow C=\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{\left(x-3\right)\left(1-x\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{2x\left(1-x\right)}{\left(x-3\right)\left(x+3\right)}\)

\(\Leftrightarrow C=\dfrac{\left(x+1\right)\left(x+3\right)-\left(x-3\right)\left(1-x\right)+2x\left(1-x\right)}{\left(x-3\right)\left(x+3\right)}\)

\(\Leftrightarrow C=\dfrac{\left(x^2+3x+x+3\right)-\left(x-x^2-3+3x\right)+\left(2x-2x^2\right)}{\left(x-3\right)\left(x+3\right)}\)

\(\Leftrightarrow C=\dfrac{x^2+3x+x+3-x+x^2+3-3x+2x-2x^2}{\left(x-3\right)\left(x+3\right)}\)

\(\Leftrightarrow C=\dfrac{2x+6}{\left(x-3\right)\left(x+3\right)}\)

\(\Leftrightarrow C=\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)

\(\Leftrightarrow C=\dfrac{2}{x-3}\)

Thay \(x=5\) vào biểu thức C ta được:

\(\dfrac{2}{5-3}=\dfrac{2}{2}=1\)

Vậy giá trị của biểu thức C tại \(x=5\) là 1

\(\text{a) }\dfrac{5x^2-3x}{5}+\dfrac{3x+1}{4}< \dfrac{x\left(2x+1\right)}{2}-\dfrac{3}{2}\\ \Leftrightarrow4\left(5x^2-3x\right)+5\left(3x+1\right)< 10x\left(2x+1\right)-15\\ \Leftrightarrow20x^2-12x+15x+5< 20x^2+10x-15\\ \Leftrightarrow20x^2+3x-20x^2-10x< -15-5\\ \Leftrightarrow-7x< -20\\ \Leftrightarrow x>\dfrac{20}{7}\)

Vậy bất phương trình có nghiệm \(x>\dfrac{20}{7}\)

\(\text{b) }\dfrac{5x-20}{3}-\dfrac{2x^2+x}{2}\ge\dfrac{x\left(1-3x\right)}{3}-\dfrac{5x}{4}\\ \Leftrightarrow4\left(5x-20\right)-6\left(2x^2+x\right)\ge4x\left(1-3x\right)-15x\\ \Leftrightarrow20x-80-12x^2-6x\ge4x-12x^2-15x\\ \Leftrightarrow-12x^2+14x+12x^2+11x\ge80\\ \Leftrightarrow25x\ge80\\ \Leftrightarrow x\ge\dfrac{16}{5}\)

Vậy bất phương trình có nghiệm \(x\ge\dfrac{16}{5}\)

\(\text{c) }\left(x+3\right)^2\le x^2-7\\ \Leftrightarrow x^2+6x+9\le x^2-7\\ \Leftrightarrow x^2+6x-x^2\le-7-9\\ \Leftrightarrow6x\le-16\\ \Leftrightarrow x\le-\dfrac{8}{3}\)

Vậy bất phương trình có nghiệm \(x\le-\dfrac{8}{3}\)

bài 1:

b,\(\dfrac{x+2}{x}=\dfrac{x^2+5x+4}{x^2+2x}+\dfrac{x}{x+2}\)(ĐKXĐ:x ≠0,x≠-2)

<=>\(\dfrac{\left(x+2\right)^2}{x\left(x+2\right)}=\dfrac{x^2+5x+4}{x\left(x+2\right)}+\dfrac{x^2}{x\left(x+2\right)}\)

=>\(x^2+4x+4=x^2+5x+4+x^2\)

<=>\(x^2-x^2-x^2+4x-5x+4-4=0\)

<=>\(-x^2-x=0< =>-x\left(x+1\right)=0< =>\left[{}\begin{matrix}x=0\left(loại\right)\\x+1=0< =>x=-1\left(nhận\right)\end{matrix}\right.\)

vậy...............

d,\(\left(x+3\right)^2-25=0< =>\left(x+3-5\right)\left(x+3+5\right)=0< =>\left(x-2\right)\left(x+8\right)=0< =>\left[{}\begin{matrix}x-2=0\\x+8=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=2\\x=-8\end{matrix}\right.\)

vậy............

bài 3:

g,\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{x^2-x-2}\)(ĐKXĐ:x khác -1,x khác 2)

<=>\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{x^2-2x+x-2}\)

<=>\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{x\left(x-2\right)+\left(x-2\right)}\)

<=>\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{\left(x+1\right)\left(x-2\right)}\)

<=>\(\dfrac{4\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}-\dfrac{2\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=\dfrac{x+3}{\left(x+1\right)\left(x-2\right)}\)

=>\(4x-8-2x-2=x+3\)

<=>\(x=13\)

vậy..............

mấy ý khác bạn làm tương tụ nhé

chúc bạn học tốt ^ ^

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

\(\Leftrightarrow9x^2-6x+1-x^3-9x^2-27x-27=8-x^3\)

\(\Leftrightarrow-x^3-33x-26-8+x^3=0\)

=>-33x=34

hay x=-34/33

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

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

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

\(\Leftrightarrow2x^2=4\)

hay \(x\in\left\{\sqrt{2};-\sqrt{2}\right\}\)

c: \(x^2-2\sqrt{3}x+3=0\)

\(\Leftrightarrow\left(x-\sqrt{3}\right)^2=0\)

hay \(x=\sqrt{3}\)

d: \(\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)-\left(x-\sqrt{2}\right)^2=0\)

\(\Leftrightarrow\left(x-\sqrt{2}\right)\left(x+\sqrt{2}-x+\sqrt{2}\right)=0\)

\(\Leftrightarrow x-\sqrt{2}=0\)

hay \(x=\sqrt{2}\)

Mk xin lỗi nha, câu c sai đề

c) (x+6)⁴ + (x+8)⁴ = 272