\(x^{11}+x^7+1=x^{11}+x^7+x^4+1-x^4\)

\(=x^7\left(x^4+1\right)+\left(x^4+1\right)-x^4=\left(x^4+1\right)\left(x^7+1\right)-x^4\)

\(=\left(\sqrt{\left(x^4+1\right)\left(x^7+1\right)}+x^2\right)\left(\sqrt{\left(x^4+1\right)\left(x^7+1\right)}-x^2\right)\)

Trần Thị Mĩ Duyên Bạn ơi nếu x âm là căn thức vô nghĩa đó !

đề sai

luôn nhỏ mà

Đặt a=x²+3x+5 

ta có \(8a^2+7a-15\)

\(=8a^2-8a+15a-15=8a\left(a-1\right)+15\left(a-1\right)\)

\(=\left(8a+15\right)\left(a-1\right)\)

Trả lại biến

\(\left(8x^2+24x+40+15\right)\left(x^2+3x+5-1\right)\)

\(=\left(8x^2+24x+55\right)\left(x^2+3x+4\right)\)

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

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

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

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

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

\(\Leftrightarrow-x=-2\)

\(\Leftrightarrow x=2\)

Vậy: phương trình có tập nghiệm: S = {2}

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

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

\(\Leftrightarrow x+1=x+2\)

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

\(\Leftrightarrow-1=0\left(vl\right)\)

Vậy pt vô n_o

b) \(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{\left(x+1\right)\left(x+2\right)}\)

\(\frac{2\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}-\frac{x+1}{\left(x+1\right)\left(x-2\right)}=\frac{3x-11}{\left(x+1\right)\left(x+2\right)}\)

\(\frac{2x-4-x-1}{\left(x+1\right)\left(x-2\right)}=\frac{3x-11}{\left(x+1\right)\left(x+2\right)}\)

\(\frac{-5}{\left(x+1\right)\left(x-2\right)}=\frac{3x-11}{\left(x+1\right)\left(x+2\right)}\)

\(\Leftrightarrow-5\left(x+2\right)=\left(3x-11\right)\left(x-2\right)\)

\(-5x+2=3x^2-11x-6x+22\)

\(3x^2-17x+22+5x-2=3x^2-12x+20=0\)

đến đây mk chịu ~

a, 

Trong 1kg = 1000g than có 960g C; 20g S 

=> nC= 80 mol; nS= 0,625 mol 

C+O2to⟶CO2C+O2⟶toCO2 

S+O2to⟶SO2S+O2⟶toSO2 

=> nO2= nS+ nC= 80,625 mol 

=> V O2= 1806l 

b,  

nC= nCO2 => mCO2= 3520g

nS= nSO2 => mSO2= 1280g

Số kg than đá chứa trong 1kg than đá đó là:

1.96%=0,96(kg)=960g

⇒nC=96012=80(mol)

Theo gt ta có PTHH: C+O2−to−>CO2 (*)

Theo (*) và gt: 80mol.....80mol..........80mol

\(\left(2x-1\right)^2=49\)

\(\Rightarrow\left(2x-1\right)^2=7^2\)

\(\Rightarrow\left(2x-1\right)^2=7\)

\(\Rightarrow\orbr{\begin{cases}2x-1=-7\\2x-1=7\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}2x=-6\\2x=8\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-3\\x=4\end{cases}}\)

a)   \(\left(2x-1\right)^2=49\)

<=> \(\orbr{\begin{cases}2x-1=7\\2x-1=-7\end{cases}}\)

<=> \(\orbr{\begin{cases}x=4\left(n\right)\\x=-3\left(n\right)\end{cases}}\)

b)    \(\left(5x-3\right)^2-\left(4x-7\right)^2=0\)

<=> \(\left(5x-3-4x+7\right)\left(5x-3+4x-7\right)=0\)

<=> \(\orbr{\begin{cases}x=-4\left(n\right)\\x=\frac{10}{9}\left(n\right)\end{cases}}\)

a)    \(A=\frac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\frac{x^2-1+x+2-x^2}{x\left(x-1\right)}\right)\)

<=> \(A=\frac{x\left(x+1\right)}{\left(x-1\right)^2}.\frac{x\left(x-1\right)}{x+1}\)

<=> \(A=\frac{x^2}{x-1}\)

b) \(|2x+1|=3\)

TH1: 2x+1=3 \(\left(x\ge\frac{-1}{2}\right)\)

    => x=1 (TM)

TH2: 2x+1=-3 \(\left(x< \frac{-1}{2}\right)\)

    => x=-2 (TM)

c)     \(A< 3\)

<=> \(\frac{x^2}{x-1}< 3\)

<=> \(\frac{x^2-3x+3}{x-1}< 0\)

 =>  \(x< 1\)

\(A=\frac{x^2+x}{x^2-2x+1}:\left(\frac{x+1}{x}-\frac{1}{1-x}+\frac{2-x^2}{x^2-x}\right)\left(x\ne0;x\ne1\right)\)

\(\Leftrightarrow A=\frac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\frac{x+1}{x}+\frac{1}{x-1}+\frac{2-x^2}{x\left(x-1\right)}\right)\)

\(\Leftrightarrow A=\frac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\frac{\left(x-1\right)\left(x+1\right)}{x\left(x-1\right)}+\frac{x}{x\left(x-1\right)}+\frac{2-x^2}{x\left(x-1\right)}\right)\)

\(\Leftrightarrow A=\frac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\frac{x^2-1}{x\left(x-1\right)}+\frac{x}{x\left(x-1\right)}+\frac{2-x^2}{x\left(x-1\right)}\right)\)

\(\Leftrightarrow A=\frac{x\left(x+1\right)}{\left(x-1\right)^2}:\frac{x^2-1+x+2-x^2}{x\left(x-1\right)}\)

\(\Leftrightarrow A=\frac{x\left(x+1\right)}{\left(x-1\right)^2}:\frac{x+1}{x\left(x-1\right)}\)

\(\Leftrightarrow A=\frac{x\left(x+1\right)}{\left(x-1\right)^2}\cdot\frac{x\left(x-1\right)}{x+1}=\frac{x^2}{x-1}\)

\(2x^3+3x^2-32x=48\)

\(2x^3+3x^2-32x-48=0\)

\(\left(2x^3+3x^2\right)-\left(32x+48\right)=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x^2-16=0\\2x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}\left(x+4\right)\left(x-4\right)=0\Leftrightarrow\orbr{\begin{cases}x=-4\\x=4\end{cases}}\\x=-\frac{3}{2}\end{cases}}}\)\(\left(x+4\right)\left(x-4\right)\left(2x+3\right)=0\)

\(\hept{\begin{cases}x+4=0\\x-4=0\\2x+3=0\end{cases}\Rightarrow\hept{\begin{cases}x=-4\\x=4\\x=-\frac{3}{2}\end{cases}}}\)

\(2x^3+3x^2-32x=48\)

\(\Leftrightarrow2x^3+3x^2-32x-48=0\)

\(\Leftrightarrow\left(2x^3-32x\right)+\left(3x^2-48\right)=0\)

\(\Leftrightarrow2x\left(x^2-16\right)+3\left(x^2-16\right)=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x-4=0;x+4=0\\2x+3=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\pm4\\x=\frac{-3}{2}\end{cases}}\)

Vậy tập nghiệm của pt là S={4;-4;-3/2}

_Học tốt_

\(Đkxđ:\hept{\begin{cases}x\ne0\\x\ne3\end{cases}}\)

\(P=\frac{x^3-3x^2+6}{x^2-3x}=\frac{x^3-3x^2}{x^2-3x}+\frac{6}{x^2-3x}=x+\frac{6}{x^2-3x}\)

Để \(P\) nguyên thì \(\Leftrightarrow x+\frac{6}{x^2-3x}\) nguyên mà \(x\in Z\)

\(\Leftrightarrow\frac{6}{x^2-3x}\in Z\) 

\(\Leftrightarrow x^2-3x\inƯ\left(6\right)\)

\(\Leftrightarrow x^2-3x=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)

Đến đây tự kẻ cái bảng ra  .............. 

Vậy \(\orbr{\begin{cases}x=1\\x=2\end{cases}}\) thì \(P\)nhận giá trị nguyên.