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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:\(x\ne\frac{3}{2};x\ne\frac{5}{3}\)
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
\(x-1=0\)
\(x=1\)
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
\(\Rightarrow\orbr{\begin{cases}x-1=1\\x-1=-1\end{cases}}\)
\(\orbr{\begin{cases}x=0\left(n\right)\\x=-2\left(n\right)\end{cases}}\)
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: bạn tự tính nhé
PT tương đương: \(\frac{5}{x-1}-\frac{5}{x-3}=\frac{2}{x+1}-\frac{2}{x-4}\)
<=>\(\frac{5x-15}{\left(x-1\right)\left(x-3\right)}-\frac{5x-5}{\left(x-1\right)\left(x-3\right)}=\frac{2x-8}{\left(x+1\right)\left(x-4\right)}-\frac{2x+2}{\left(x+1\right)\left(x-4\right)}\)
<=>\(\frac{-10}{\left(x-1\right)\left(x-3\right)}=\frac{-10}{\left(x+1\right)\left(x-4\right)}\)
<=>\(\frac{1}{\left(x-1\right)\left(x-3\right)}=\frac{1}{\left(x+1\right)\left(x-4\right)}\)
<=>\(\frac{\left(x+1\right)\left(x-4\right)}{\left(x-1\right)\left(x-3\right)\left(x+1\right)\left(x-4\right)}=\frac{\left(x-1\right)\left(x-3\right)}{\left(x-1\right)\left(x-3\right)\left(x+1\right)\left(x-4\right)}\)
=>\(\left(x+1\right)\left(x-4\right)=\left(x-1\right)\left(x-3\right)\)
Còn lại bạn từ làm nhé:)
ĐKXĐ: x∉{1;3}
Ta có: \(\frac{x-5}{x-1}+\frac{2}{x-3}=1\)
\(\Leftrightarrow\frac{\left(x-5\right)\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}+\frac{2\left(x-1\right)}{\left(x-3\right)\left(x-1\right)}=\frac{\left(x-1\right)\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}\)
Suy ra: \(x^2-8x+15+2x-2=x^2-4x+3\)
⇔\(x^2-6x+13-x^2+4x-3=0\)
\(\Leftrightarrow-2x+10=0\)
⇔\(-2x=-10\)
hay x=5(tm)
Vậy: x=5
\(PT< =>\frac{\left(x-5\right)\left(x-3\right)+2\left(x-1\right)}{\left(x-1\right)\left(x-3\right)}=1\)
<=> \(\frac{x^2-8x+15+2x-2}{\left(x-1\right)\left(x-3\right)}=1\)
<=> \(x^2-6x+13=x^2-4x+3\)
<=> \(x^2-6x+13-x^2+4x-3=0\)
<=> \(-2x+10=0\)
<=> x = 5 (TMDK)