Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1a.
ĐKXĐ: \(x\ne\left\{1;3\right\}\)
\(\Leftrightarrow\dfrac{6}{x-1}=\dfrac{4}{x-3}+\dfrac{4}{x-3}\)
\(\Leftrightarrow\dfrac{3}{x-1}=\dfrac{4}{x-3}\Leftrightarrow3\left(x-3\right)=4\left(x-1\right)\)
\(\Leftrightarrow3x-9=4x-4\Rightarrow x=-5\)
b.
ĐKXĐ: \(x\ne\left\{-1;2\right\}\)
\(\Leftrightarrow\dfrac{5}{x+1}=\dfrac{3}{2-x}+\dfrac{1}{2-x}\)
\(\Leftrightarrow\dfrac{5}{x+1}=\dfrac{4}{2-x}\Leftrightarrow5\left(2-x\right)=4\left(x+1\right)\)
\(\Leftrightarrow10-2x=4x+4\Leftrightarrow6x=6\Rightarrow x=1\)
1c.
ĐKXĐ: \(x\ne\left\{2;5\right\}\)
\(\Leftrightarrow\dfrac{3x\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}-\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x-5\right)}=\dfrac{-3x}{\left(x-2\right)\left(x-5\right)}\)
\(\Leftrightarrow3x\left(x-5\right)-x\left(x-2\right)=-3x\)
\(\Leftrightarrow2x^2-10x=0\Leftrightarrow2x\left(x-5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=5\left(loại\right)\end{matrix}\right.\)
2a.
\(\Leftrightarrow-4x^2-5x+6=x^2+4x+4\)
\(\Leftrightarrow5x^2+9x-2=0\Rightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{5}\end{matrix}\right.\)
2b.
\(2x^2-6x+1=0\Rightarrow x=\dfrac{3\pm\sqrt{7}}{2}\)
\(a,\dfrac{3}{2x-1}+1=\dfrac{2x-1}{2x+1};ĐKXĐ:x\ne\pm\dfrac{1}{2}\\ \Leftrightarrow\dfrac{3}{2x-1}-\dfrac{2x-1}{2x+1}+1=0\\ \Leftrightarrow\dfrac{3\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)}-\dfrac{\left(2x-1\right)\left(2x-1\right)}{\left(2x+1\right)\left(2x-1\right)}+\dfrac{\left(2x-1\right)\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)}=0\\ \Rightarrow3\left(2x+1\right)-\left(2x-1\right)^2+\left(2x-1\right)\left(2x+1\right)=0\\ \Leftrightarrow6x+3-\left(4x^2-4x+1\right)+\left(4x^2-1\right)=0\\ \Leftrightarrow6x+3-4x^2+4x-1+4x^2-1=0\\ \Leftrightarrow10x+1=0\\ \Leftrightarrow10x=-1\\ \Leftrightarrow x=-\dfrac{1}{10}\)
Vậy \(x\in\left\{-\dfrac{1}{10}\right\}\)
1/ ĐKXĐ : \(x\ne1\)
\(\dfrac{7x-3}{x-1}=\dfrac{2}{3}\)
\(\Leftrightarrow21x-9=2x-2\)
\(\Leftrightarrow19x=7\Leftrightarrow x=\dfrac{7}{19}\left(tm\right)\)
Vậy...
b/ \(\dfrac{2\left(3-7x\right)}{1+x}=\dfrac{1}{2}\) ĐKXĐ : \(x\ne-1\)
\(\Leftrightarrow12-28x=1+x\)
\(\Leftrightarrow11=29x\Leftrightarrow x=\dfrac{11}{29}\) \(\left(tm\right)\)
Vậy....
c/ ĐKXĐ : \(x\ne0\)
\(\dfrac{x^2-6}{x}=x+\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{x^2-6}{x}=\dfrac{2x+3}{2}\)
\(\Leftrightarrow2x^2-12=2x^2+3x\)
\(\Leftrightarrow3x=-12\Leftrightarrow x=-4\) \(\left(tm\right)\)
Vậy...
4/ ĐKXĐ : \(x\ne-\dfrac{2}{3}\)
\(\dfrac{5}{3x+2}=2x-1\)
\(\Leftrightarrow\left(2x-1\right)\left(3x+2\right)=5\)
\(\Leftrightarrow6x^2+4x-3x-2=5\)
\(\Leftrightarrow6x^2+x-7=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{6}\\x=1\end{matrix}\right.\)
Vậy....
5,6 Tương tự nhé !
1)ĐKXĐ: \(x\ne1\)
Ta có: \(\dfrac{7x-3}{x-1}=\dfrac{2}{3}\)
\(\Leftrightarrow3\left(7x-3\right)=2\left(x-1\right)\)
\(\Leftrightarrow21x-9=2x-2\)
\(\Leftrightarrow21x-9-2x+2=0\)
\(\Leftrightarrow19x-7=0\)
\(\Leftrightarrow19x=7\)
\(\Leftrightarrow x=\dfrac{7}{19}\)(nhận)
Vậy: \(S=\left\{\dfrac{7}{19}\right\}\)
2) ĐKXĐ: \(x\ne-1\)
Ta có: \(\dfrac{2\left(3-7x\right)}{1+x}=\dfrac{1}{2}\)
\(\Leftrightarrow4\left(3-7x\right)=x+1\)
\(\Leftrightarrow12-28x-x-1=0\)
\(\Leftrightarrow-29x+11=0\)
\(\Leftrightarrow-29x=-11\)
\(\Leftrightarrow x=\dfrac{11}{29}\)
Vậy: \(S=\left\{\dfrac{11}{29}\right\}\)
3) ĐKXĐ: \(x\ne0\)
Ta có: \(\dfrac{x^2-6}{x}=x+\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{x^2-6}{x}=\dfrac{2x+3}{2}\)
\(\Leftrightarrow2\left(x^2-6\right)=x\left(2x+3\right)\)
\(\Leftrightarrow2x^2-12=2x^2+6x\)
\(\Leftrightarrow2x^2-12-2x^2-6x=0\)
\(\Leftrightarrow-6x-12=0\)
\(\Leftrightarrow-6x=12\)
\(\Leftrightarrow x=-2\)
Vậy: S={-2}
1, bạn xem lại đề
2, 15(x-3) + 8x-21 = 12(x+1) +120
<=> 23x - 66 = 12x + 132
<=> 11x = 198 <=> x = 198/11
3, 10(3x+1) + 5 - 100 = 8(3x-1) - 6x - 4
<=> 30x + 10 - 95 = 18x -12
<=> 12x = 73 <=> x = 73/12
a) Ta có: \(\dfrac{2x+1}{6}-\dfrac{x-2}{4}=\dfrac{3-2x}{3}-x\)
\(\Leftrightarrow\dfrac{2\left(2x+1\right)}{12}-\dfrac{3\left(x-2\right)}{12}=\dfrac{4\left(3-2x\right)}{12}-\dfrac{12x}{12}\)
\(\Leftrightarrow4x+2-3x+6=12-8x-12x\)
\(\Leftrightarrow x+8-12+20x=0\)
\(\Leftrightarrow21x-4=0\)
\(\Leftrightarrow21x=4\)
\(\Leftrightarrow x=\dfrac{4}{21}\)
Vậy: \(S=\left\{\dfrac{4}{21}\right\}\)
Hình như em viết công thức bị lỗi rồi. Em cần chỉnh sửa lại để được hỗ trợ tốt hơn!
b: \(\Leftrightarrow\dfrac{7x+10}{x+1}\left(x^2-x-2-2x^2+3x+5\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(-x^2+2x+3\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(x^2-2x-3\right)=0\)
=>(7x+10)(x-3)=0
hay \(x\in\left\{-\dfrac{10}{7};3\right\}\)
d: \(\Leftrightarrow\dfrac{13}{2x^2+7x-6x-21}+\dfrac{1}{2x+7}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\dfrac{13}{\left(2x+7\right)\left(x-3\right)}+\dfrac{1}{\left(2x+7\right)}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow26x+91+x^2-9-12x-14=0\)
\(\Leftrightarrow x^2+14x+68=0\)
hay \(x\in\varnothing\)
Câu 1:
=>15(2x+1)-8(3x-1)=100
=>30x+15-24x+8=100
=>6x+23=100
hay x=77/6
Câu 2:
=>2(5x-3)+12-3(7x-1)=x+2
=>10x-6+12-21x+3-x-2=0
=>-12x=-7
hay x=7/12
Câu 3:
\(\Leftrightarrow2\left(x^2-1\right)+3\left(x+1\right)=2\left(x^2-4x+4\right)\)
\(\Leftrightarrow2x^2-2+3x+3-2x^2+8x-8=0\)
=>11x-7=0
hay x=-7/11
Câu 4:
(x - 4)^3/6 + 1 = x(x + 1)/2 - (x - 5)(x + 5)/3
<=> (x - 4)^3 + 6/6 = x^2 + x/2 - x^2 - 25/3
<=> (x - 4)^3 + 6/6 = 3x^2 + 3x - 2x^2 + 50/6
<=> (x - 4)^3 + 6 = 3x^2 + 3x - 2x^2 + 50
<=> x^3 - 12x^2 + 48x - 58 = x^2 + 3x + 50
<=> x^3 -13x^2 + 45x - 108 = 0
Đến đây bạn bấm máy nhẩm nghiệm là ra nhé
Câu 5:
3(x + 2)^3/5 - (x - 1)^2/10 = (x - 3)(x + 3)/2
<=> 6(x + 2)^3 - (x - 1)^2/10 = 5(x^2 - 9)/10
<=> 6(x + 2)^3 - (x - 1)^2 = 5(x^2 - 9)
<=> 6x^3 + 36x^2 + 72x + 48 - x^2 + 2x - 1 - 5x^2 + 45 = 0
<=> 6x^3 + 30x^2 + 74x + 92 = 0
Đến đây bạn bấm máy nhẩm nghiệm như câu 4 nhé
Bạn chia nhỏ các phần ra nhé.
uh mk biết lần sau mk rút kinh nghiệm