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a) \(\dfrac{x+1}{4}-\dfrac{5+2x}{8}=\dfrac{3-4x}{2}\)
⇔\(\dfrac{2\left(x+1\right)}{8}-\dfrac{5+2x}{8}=\dfrac{4\left(3-4x\right)}{8}\)
⇔ 2x + 2 - 5 - 2x = 12 -16x
⇔ 16x = 15
⇔ x = 15/16
b) \(\dfrac{4-3x}{5}-\dfrac{4-x}{10}=\dfrac{x+2}{2}\)
⇔\(\dfrac{2\left(4-3x\right)}{10}-\dfrac{4-x}{10}=\dfrac{5\left(x+2\right)}{10}\)
⇔ 8 - 6x - 4 + x = 5x + 10
⇔ 10x = -6
⇔ x = -6/10
Câu 1:
x + 1/4 - 5 + 2x/8 = 3 - 4x/2
<=> 2x + 2/8 - 5 + 2x/8 = 12 - 16x/8
<=> 2x + 2 - 5 - 2x = 12 - 16x
<=> -3 = 12 - 16x <=> 15 = 16x <=> x = 15/16
Câu 2:
4 - 3x/5 - 4 - x/10 = x + 2/2
<=> 8 - 6x/10 - 4 - x/10 = 5x + 10/10
<=> 8 - 6x - 4 + x = 5x + 10
<=> 4 - 5x = 5x + 10
<=> 4 = 10x + 10 <=> 10x = -6 <=> x = -3/5
a: \(P=\dfrac{8+5x-2x-8}{x\left(x+4\right)}=\dfrac{3x}{x\left(x+4\right)}=\dfrac{3}{x+4}\)
b: Khi x=1/2 thì P=3/(1/2+4)=3:9/2=3*2/9=6/9=2/3
ý bạn là tìm x hay sao?
\(a,\Leftrightarrow\dfrac{\left(x-2\right)\left(x+1\right)}{x+1}=\dfrac{x^2-3x-2}{x-1}\left(x\ne\pm1\right)\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=x^2-3x-2\\ \Leftrightarrow x^2-3x+2=x^2-3x-2\\ \Leftrightarrow2=-2\Leftrightarrow x\in\varnothing\\ b,\Leftrightarrow\dfrac{\left(x+2\right)\left(x^2-2x+4\right)}{x^2-2x+4}=x+2\\ \Leftrightarrow x+2=x+2\\ \Leftrightarrow x\in R\)
alo cho tui hỏi bạn có phải Dương Ngọc Lan Hương Trường THCS Minh Thuận 3 k dọ
a: ĐKXĐ: x<>0; x<>-1
PT =>x+1-2x=3
=>1-x=3
=>x=-2(nhận)
b: Sửa đề: \(\dfrac{1}{2x-3}-\dfrac{3}{x\left(2x-3\right)}=\dfrac{5}{x}\)
=>x-3=5(2x-3)
=>10x-15=x-3
=>9x=12
=>x=4/3(nhận)
c: ĐKXĐ: x<>0; x<>2
PT =>x(x+2)-x+2=2
=>x^2+2x-x=0
=>x(x+1)=0
=>x=-1
A.
\(\Leftrightarrow\) 9x - 2x - 6 = 3x + 1
\(\Leftrightarrow\) 4x = 7
\(\Leftrightarrow\) x = \(\dfrac{7}{4}\)
B.
\(\Leftrightarrow\dfrac{5\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{4\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{x-13}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow\) 5x + 15 - 4x +12 = x - 13
\(\Leftrightarrow\) 0x = -40 ( phương trình vô nghiệm)
C.
\(\Leftrightarrow\) 7x + 8 \(\ge\) 3x -3
\(\Leftrightarrow\) 4x \(\ge\) - 11
\(\Leftrightarrow\)\(x\ge\dfrac{-11}{4}\)
Bạn ơi mik ra \(\dfrac{x^3+45x-54}{12\left(x-3\right)\left(x+3\right)}\) có đúng không bạn?
Mình rút chx hết bạn bạn gửi cách làm bạn qua mình tham khảo đc k ạ?
\(\Leftrightarrow6x-3-4x+20< =4x-1+24\)
=>2x+17-4x-23<=0
=>-2x-6<=0
=>-2x<=6
hay x>=-3
\(\Leftrightarrow\dfrac{6x-3-4x+20-4x+1-24}{12}\le0\)
\(\Rightarrow\left\{{}\begin{matrix}-2x-6< 0\\-2x-6\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>-3\\x\ne-3\end{matrix}\right.\)
Câu 1:
\(\left(x-2\right)\left(x^2+2x+4\right)+25x=x\left(x+5\right)\left(x-5\right)+8\)
\(\Leftrightarrow x^3-8+25x=x\left(x^2-25\right)+8\)
\(\Leftrightarrow x^3-8+25x=x^3-25x+8\)
\(\Leftrightarrow x^3-8+25x-x^3+25x-8=0\)
\(\Leftrightarrow50x-16=0\)
\(\Leftrightarrow50x=16\)
\(\Leftrightarrow x=\dfrac{8}{25}\)
Câu 2 :
\(\dfrac{x+5}{4}+\dfrac{3+2x}{3}=\dfrac{6x-1}{3}-\dfrac{1-2x}{12}\)
<=> \(\dfrac{3\left(x+5\right)}{12}+\dfrac{4\left(3+2x\right)}{12}=\dfrac{4\left(6x-1\right)}{12}-\dfrac{1-2x}{12}\)
<=>\(\dfrac{3x+15+12+8x}{12}=\dfrac{24x-4-1+2x}{12}\)
<=> 3x + 15 + 12 + 8x = 24x - 4 - 1 +2x
<=> 11x+27 = 26x -5
<=> ( 26x - 5 ) - ( 11x + 27 ) = 0
<=> 15x - 32 = 0
<=> 15x = 32
<=> x = \(\dfrac{32}{15}\)
\(\Leftrightarrow\dfrac{2\left(x+2\right)-8x+24-1+x}{8}=0\)
\(\Leftrightarrow2x+4-8x+24-1+x=0\)
\(\Leftrightarrow-5x+27=0\)
\(\Leftrightarrow x=\dfrac{27}{5}\)