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

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

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

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

\(\Leftrightarrow6x^2+2=6x^2+6x+6\)

\(\Leftrightarrow6x^2+2-6x^2-6x-6=0\)

\(\Leftrightarrow-6x-4=0\)

\(\Leftrightarrow-6x=4\)

\(\Leftrightarrow x=\frac{-4}{6}=\frac{-2}{3}\)

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

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

\(\Leftrightarrow\left(\frac{2x+12}{3}\right)\left(x+3\right)\left(\frac{x-2}{2}\right)=0\)

\(\Leftrightarrow\frac{2\left(x+6\right)\left(x+3\right)}{3}.\frac{\left(x-2\right)}{2}=0\)

\(\Leftrightarrow\frac{\left(x+6\right)\left(x+3\right)\left(x-2\right)}{3}=0\)

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

\(\Leftrightarrow\hept{\begin{cases}x+6=0\\x+3=0\\x-2=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=-6\\x=-3\\x=2\end{cases}}\)

Bạn đổi dấu ngoặc nhọn thành ngoặc vuông giúp mình nhé

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

\(\Leftrightarrow2\left(\frac{x}{3}+2\right)\left(x+4\right)\left(\frac{x}{2}-1\right)=0\)

\(\Leftrightarrow2\left(x+6\right)\left(x+4\right)\left(x-2\right)=0\)

\(\Leftrightarrow x=-6;x=-4;x=2\)

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

\(\Leftrightarrow\left(x^2-6x+9\right)-\left(2x-2\right)=x\left(x^2-4x+4\right)-5x^2\)

\(\Leftrightarrow x^2-6x+9-2x+2=x^3-4x^2+4x-5x^2\)

\(\Leftrightarrow x^2-8x+11=x^3-9x^2+4x\)

Đưa về bậc 3 giải nốt

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

\(\Leftrightarrow\left(x-3\right)^3-2\left(x-1\right)-x\left(x-2\right)^2+5x^2=0\)

\(\Rightarrow x^3-9x^2+27x-27-2x-x^3+4x^2-4x+5x^2=0\)

\(\Rightarrow21x=25\Leftrightarrow x=\frac{25}{21}\)

Câu 1:

a) \(\left(x^2+y^2-36\right)^2-4x^2y^2\)

\(=\left(x^2+y^2-36\right)^2-\left(2xy\right)^2\)

\(=\left(x^2+y^2+2xy-36\right)\left(x^2+y^2-2xy-36\right)\)

\(=\left[\left(x+y\right)^2-36\right]\left[\left(x-y\right)^2-36\right]\)

\(=\left(x+y+6\right)\left(x+y-6\right)\left(x-y+6\right)\left(x-y-6\right)\)

b) \(\left(x^2+x\right)^2-5\left(x^2+x\right)+6\)

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

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

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

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

1) a) (x² + y² - 36)² - 4x²y² 

= (x² + y² - 36 - 2xy)(x² + y² - 36 + 2xy)

= [(x - y)² - 36][(x + y)² - 36]

= (x - y - 6)(x - y  + 6)(x + y + 6)(x + y - 6)

b) (x² + x)² - 5(x² + x) + 6

= (x² + x)² - 2(x² + x) - 3(x² + x) + 6

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

= (x² + x - 3)(x² + 2x - x - 2)

=  (x² + x - 3)(x - 1)(x + 2)

2) Đặt tính là đc

a) Rút gọn :

ĐKXĐ : \(x\ne4,x\ne3\)

Ta có : \(Q=\frac{12x-45}{x^2-7x+12}-\frac{x+5}{x-4}+\frac{2x-3}{3-x}\)

\(=\frac{3\left(4x-15\right)}{\left(x-4\right)\left(x-3\right)}-\frac{\left(x+5\right)\left(x-3\right)}{\left(x-4\right)\left(x-3\right)}-\frac{\left(2x-3\right)\left(x-4\right)}{\left(x-4\right)\left(x-3\right)}\)

\(=\frac{12x-45-x^2-2x+15-2x^2+11x-12}{\left(x-4\right)\left(x-3\right)}\)

\(=\frac{-3x^2+21x-42}{\left(x-4\right)\left(x-3\right)}\)

... Chắc tui rút gọn sai òi :))

đkxđ \(x\ne2\)

\(\Leftrightarrow mx-2m-5x+10+2m+5=0\)

\(\Leftrightarrow mx-5x+15=0\)

\(\Leftrightarrow x\left(m-5\right)=-15\)(1)

Biện luận:

+ Nếu \(m=5\)pt (1) 0x = -15 (vô nghiệm)

+ Nếu \(m\ne5\)pt (1) có nghiệm    (2)

Kết hợp điều kiện \(\Rightarrow x=-\frac{15}{m-5}\ne2\)

                                \(\Leftrightarrow-15\ne2m-10\)

                                \(\Leftrightarrow-2m\ne5\)

                                \(\Leftrightarrow m\ne-\frac{5}{2}\)₍₃₎

Kết luận (2)(3) m = 5 thì pt vô nghiệm

                         \(m\ne5;\frac{5}{2}\)thì pt có nghiệm \(x=-\frac{15}{m-5}\)

 \(a,2H_2+O_2\rightarrow2H_2O\)

\(2CO+O_2\rightarrow2CO_2\)

Gọi \(a\) là số mol \(H_2\),\(b\) là số mol \(CO\)

\(n_{O_2}=\frac{89,6}{22,4}=4\left(mol\right)\)

Ta có:\(\hept{\begin{cases}2a+28b=68\\0,5+a+0,5b=4\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}a=6\\b=2\end{cases}}\)

\(b,m_{H_2}=6.2=12g\)

\(m_{CO}=2.28=56\left(g\right)\)

\(c,\%V_{H_2}=\frac{6}{8}.100\%=75\%\)

\(\%V_{CO}=100-75=25\%\)