1. Trong các phương trình sau, phương trình bậc nhất 1 ẩn là

A. 2/x - 7=0; B. |7x+5)-1=0; C. 8x-9=0

2. điều kiện xác định của phương trình

\(\frac{4}{2x-3}=\frac{7}{3x-5}\)là

A. x khác 3/2. B. x khác5/3; C. x khác 3/2 hoặc 5/3; D. x khác 3/2 và 5/3

1.Pt bậc nhất 1 ẩn:\(8x-9=0\)

2.ĐKXĐ:\(x\ne\frac{3}{2};x\ne\frac{5}{3}\)

đề sai k bạn

a) \(\frac{x-1}{2}+\frac{x-2}{3}+\frac{x-3}{4}=\frac{x-4}{5}+\frac{x-5}{6}\)

\(\left(\frac{x-1}{2}+1\right)+\left(\frac{x-2}{3}+3\right)+\left(\frac{x-3}{4}+1\right)=\left(\frac{x-4}{5}+1\right)+\left(\frac{x-5}{6}+1\right)\)

\(\frac{x-1}{2}+\frac{x-1}{3}+\frac{x-1}{4}=\frac{x-1}{5}+\frac{x-1}{6}\)

\(\left(x-1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)\)=0

\(x-1=0\)

\(x=1\)

có pt mô mà giải

ĐKXĐ \(x\ge0,x\ne1\)

Sai điều kiện hay sao á 

Điều kiện là x - 1 khác 0 

x khác 1 

\(E=\frac{x^2}{x-1}\)   

\(=\frac{x^2-1+1}{x-1}\)   

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

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

\(=x+1+\frac{1}{x-1}\)   

Để thỏa đề thì 1 phải chia hết cho x - 1 

x - 1 là ước của 1 

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

\(\orbr{\begin{cases}x=0\left(n\right)\\x=-2\left(n\right)\end{cases}}\)

ĐK đúng đó bạn

Câu 1:

a)\(x^2-4+\left(x-2\right)\left(2x+1\right)=0\)

\(\Rightarrow x^2-4+2x^2+x-4x-2=0\)

\(\Rightarrow3x^2-3x-6=0\)

\(\Rightarrow x^2-x-2=0\)(Vì nhân tử chung là 3 thì ra bằng 0)

\(\Rightarrow x^2-2x+x-2=0\)

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

\(\Rightarrow\orbr{\begin{cases}x+1=0\\x-2=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)

         Vậy x=-1;2

Câu 2:

a)\(ĐKXĐ:X\ne1;X\ne-1;X\ne-2;\)

b)\(\frac{x+1}{x-1}-\frac{x-1}{x+2}=\frac{3}{x^2-1}\)(\(ĐKXĐ:X\ne1;X\ne-1;X\ne-2;\))

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

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

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

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

\(\Rightarrow\left(x+1\right)\left[5x+1\right]=3x+6\)

\(\Rightarrow5x^2+6x+1-3x-6=0\)

\(\Rightarrow5x^2+3x-5=0\)

\(\Rightarrow x=0,745\left(TM\right)\)

a)Ta có:\(1-2x=\frac{-7x-11}{5}\)

\(\Rightarrow\frac{5-10x}{5}=\frac{-7x-11}{5}\)

\(\Rightarrow5-10x=-7x-11\)

\(\Rightarrow5-10x+7x+11=0\)

\(\Rightarrow16-3x=0\)

\(\Rightarrow x=\frac{16}{3}\)

A . 3x + 2(x + 1) = 6x - 7

<=> 3x + 2x + 2 = 6x -7

<=> 5x - 6x = -7 - 2

<=> -x = -9

<=> x =9

B . \(\frac{x+3}{5}\).< \(\frac{5-x}{3}\)

=> 3(x +3) < 5(5 -x)

<=> 3x+9 < 25 - 5x

<=> 3x + 5x < 25 - 9

<=> 8x < 16

<=> x < 2

C . \(\frac{5}{x+1}\)+ \(\frac{2x}{x^2-3x-4}\)=\(\frac{2}{x-4}\)

<=> \(\frac{5}{x+1}\)+ \(\frac{2x}{x^2+x-4x-4_{ }}\)= \(\frac{2}{x-4}\)

<=> \(\frac{5}{x+1}\)+ \(\frac{2x}{\left(x+1\right)\left(x-4\right)}\)= \(\frac{2}{x-4}\)

<=> 5(x - 4) + 2x = 2(x +1)

<=> 5x - 20 + 2x = 2x + 2

<=>7x - 2x = 2 + 20

<=> 5x = 22

<=> x =\(\frac{22}{5}\)

\(a,\Leftrightarrow5\left(x-2\right)-15x\le9+10\left(x+1\right)\)

\(\Leftrightarrow5x-10-15x\le9+10x+10\)

\(\Leftrightarrow-20x\le29\)

\(\Leftrightarrow x\ge-1,45\)

Vậy ...........

\(b,\Rightarrow\left(x+2\right)-3\left(x-3\right)=5\left(x-2\right)\)

\(\Leftrightarrow x+2-3x+9-5x+10=0\)

\(\Leftrightarrow-7x+21=0\)

\(\Leftrightarrow x=3\)

Vậy ..............

 \(\frac{x-2}{6}-\frac{x}{2}\le\frac{3}{10}+\frac{x+1}{3}\Leftrightarrow\frac{5\left(x-2\right)}{30}-\frac{15x}{30}\le\frac{9}{30}+\frac{10\left(x+1\right)}{30}\)

\(\Leftrightarrow5x-10-15x-9-10x-10\le0\) 

 \(\Leftrightarrow-20x-29\le0\Leftrightarrow\left(-20x\right)\cdot\frac{-1}{20}\ge29\cdot-\frac{1}{20}\)

 \(\Leftrightarrow x\ge-\frac{29}{20}\)

ĐKXĐ : \(\hept{\begin{cases}x\ne2\\x\ne3\end{cases}}\)

\(\frac{x+2}{x^2-5x+6}-\frac{3}{x-2}=\frac{5}{x-3}\) 

 \(\Rightarrow\frac{x+2}{x-2x-3x+6}-\frac{3}{x-2}=\frac{5}{x-3}\)

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

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

 \(\Rightarrow x+2-3x+9-5x+10=0\)

\(\Leftrightarrow-7x+21=0\Leftrightarrow x=3\) (nhân)

 tập nghiệm của phương trình là S= 3