Có phải đề sau không ?

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

Nếu là đề trên thì cách giải như sau

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

\(\Rightarrow3x+1+2x-x+1=0\)

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

\(\Rightarrow4x+2=0\)

\(\Rightarrow4x=-2\Rightarrow x=\frac{-2}{4}=\frac{-1}{2}\)

Vậy \(x=\frac{-1}{2}\)

ko phải là x mũ 3 vs lại x mũ 2

1/(a+b-x) = 1/a + 1/b +1/x

\(\Leftrightarrow\frac{1}{a+b-c}-\frac{1}{x}=\frac{1}{a}+\frac{1}{b}\)

\(\Leftrightarrow\frac{2x-a-b}{\left(a+b-\right)x}=\frac{a+b}{ab}\)

\(\Rightarrow\left(2x-a-b\right)ab=\left(a+b\right)\left(a+b-x\right)x\)

<=>(a+b)x²-(a+b)²x-ab(a+b)=0

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

\(16x^4\)\(-1\)

\(16x^4\)\(+8x^3\)\(+4x^2\)\(+2x-8x^3\)\(-4x^2\)\(-2x-1\)

\(2\left(8x^4+4x^3+2x^2+x\right)\)\(-1\left(8x^3+4x^2+2x+1\right)\)

\(2x\left(8x^3+4x^2+2x+1\right)\)\(-1\left(8x^3+4x^2+2x+1\right)\)

\(\left(2x-1\right)\)\(\left(8x^3+4x^2+2x+1\right)\)

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

Làm hơi tắt mong bạn thông cảm

a)  6\(⋮\)x-1<=>x-1\(\inướccủa6\)

<=> Ư(6)=(1;2;3;6)

x-1=1=>x=2

x-1=2=>x=3

x-1=3=>x=4

x-1=6=>x=7

14\(⋮2x+3\Rightarrow2x+3\inƯ\left(14\right)\)

Ư(14)=(1;2;7;14)

2x+3=1=>x=-1

2x+3=2=>x=-1/2

2x+3=7=>x=2

2x+3=14=>x=11/2

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

\(ĐKXĐ:x\ne\pm2\)

Đặt \(\frac{x-1}{x+2}=a;\frac{x+1}{x-2}=b\)

=> Phương trình (1) <=> \(a^2-4ab+3b^2=0\)

\(\Leftrightarrow a^2-3ab-ab+3b^2=0\)

\(\Leftrightarrow a\left(a-b\right)-3b\left(a-b\right)=0\)

\(\Leftrightarrow\left(a-3b\right)\left(a-b\right)=0\)

\(\Leftrightarrow\left(a-3b\right)\left(a-b\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a-3b=0\\a-b=0\end{cases}\Leftrightarrow\orbr{\begin{cases}a=3b\\a=b\end{cases}}}\)

=>  \(b=0;a=0\)

Bạn cùng trường :">

a) = (x + 3)² - y² = (x + 3 - y)(x + 3 + y)

b) = x²(x - 3) -4(x - 3) = (x - 3)(x² - 4) = (x - 3)(x - 2)(x + 2)

c) = 3x(x - y) - 5(x - y) = (x - y)(3x - y)

d) Nhầm đề. tui sửa lại x³ + y³ + 2x² - 2xy + 2y²

= x³ + y³ + 2(x² - xy + y²) = (x + y)(x² - xy + y²) + 2(x² - xy + y²) = (x² - xy + y²)(x + y + 2)

e) = x⁴ - x³ - x³ + x² - x² + x + x - 1 = x³(x - 1) - x²(x - 1) - x(x - 1) + x - 1 = (x - 1)(x³ - x² - x + 1) = (x - 1)(x - 1)(x² - 1) = (x - 1)³(x + 1)

f) = x³ - 3x² - x² + 3x + 9x - 27 = x²(x - 3) - x(x - 3) + 9(x - 3) = (x-3)(x² - x + 9)

g) chắc là 3xyz 

= x²y + xy² + y²z + yz² + x²z + xz² + 3xyz = x²y + xy² + xyz + y²z + yz² + xyz + x²z + xz² + xyz = (x + y + z)(xy + yz + xz)

h) = 2³ -(3x)³ = (2 - 3x)(4 + 6x + 9x²)

i) = (x + y - x + y)(x + y + x - y) = 2y*2x = 4xy

k) = (x³ - y³)(x³ + y³) = (x - y)(x² + xy +y2)(x + y)(x² - xy +y2).