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\(d,\frac{10x+3}{8}=\frac{7-8x}{12}\)
\(\left(10x+3\right):8=\left(7-8x\right):12\)
\(\left(10x+3\right).\frac{1}{8}=\left(7-8x\right).\frac{1}{12}\)
\(\frac{5}{4}x+\frac{3}{8}=\frac{7}{12}-\frac{8}{12}x\)
\(\frac{5}{4}x+\frac{8}{12}x=\frac{7}{12}-\frac{3}{8}\)
\(\frac{23}{12}x=\frac{5}{24}\)
\(x=\frac{5}{46}\)
E mới lớp 6 nên giải sai thì thông cảm ạ UwU
\(b,\frac{x}{10}-\left(\frac{x}{30}+\frac{2x}{45}\right)=\frac{4}{5}\)
\(< =>\frac{9x}{90}-\frac{7x}{90}=\frac{4}{5}\)
\(< =>\frac{x}{45}=\frac{32}{45}\)
\(< =>x=32\)
\(d,\frac{10x+3}{8}=\frac{7-8x}{12}\)
\(< =>\left(10x+3\right).12=\left(7-8x\right).8\)
\(< =>120x+36=56-64x\)
\(< =>184x=56-36=20\)
\(< =>x=\frac{20}{184}=\frac{5}{46}\)
a) \(\frac{1-x}{x+1}+3=\frac{2x+3}{x+1}\)
<=> 1 - x + 3(x + 1) = 2x + 3
<=> 1 - x + 3x + 3 = 2x + 3
<=> 1 - x + 3x + 3 - 2x = 3
<=> 4 = 3 (vô lý)
=> pt vô nghiệm
b) ĐKXĐ: \(x\ne1;x\ne2\)
\(\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(2-x\right)}\)
<=> (x - 2)(2 - x) - 5(x + 1)(2 - x) = 15(x - 2)
<=> 2x - x2 - 4 + 2x - 5x - 5x2 + 10 = 15x - 30
<=> -x + 4x2 - 14 = 15x - 30
<=> x - 4x2 + 14 = 15x - 30
<=> x - 4x2 + 14 + 15x - 30 = 0
<=> 16x - 4x2 - 16 = 0
<=> 4(4x - x2 - 4) = 0
<=> -x2 + 4x - 4 = 0
<=> x2 - 4x + 4 = 0
<=> (x - 2)2 = 0
<=> x - 2 = 0
<=> x = 2 (ktm)
=> pt vô nghiệm
c) xem bài 4 ở đây: Câu hỏi của gjfkm
d) ĐKXĐ: \(x\ne1;x\ne2;x\ne3\)
\(\frac{x+4}{x^2-3x+2}+\frac{x+1}{x^2-4x+3}=\frac{2x+5}{x^2-4x+3}\)
<=> \(\frac{x+4}{\left(x-1\right)\left(x-2\right)}+\frac{x+1}{\left(x-1\right)\left(x-3\right)}=\frac{2x+5}{\left(x-1\right)\left(x-3\right)}\)
<=> (x + 4)(x - 3) + (x + 1)(x - 2) = (2x + 5)(x - 2)
<=> x2 - 3x + 4x - 12 + x2 - 2x + x - 2 = 2x2 - 4x + 5x - 10
<=> 2x2 - 14 = 2x2 + x - 10
<=> 2x2 - 14 - 2x2 = x - 10
<=> -14 = x - 10
<=> -14 + 10 = x
<=> -4 = x
<=> x = -4
a) \(\frac{15x-10}{x^2+3}=0\)
<=> 15x - 10 = 0
<=> 5(3x - 2) = 0
<=> 3x - 2 = 0
<=> 3x = 2
<=> x = 2/3
b) ĐKXĐ: \(x\ne1;x\ne-3\)
<=>\(\frac{3x-1}{x-1}-\frac{2x+5}{x+3}-\frac{8}{x^2+2x-3}=0\)
<=> \(\frac{3x-1}{x-1}-\frac{2x+5}{x+3}-\frac{8}{\left(x-1\right)\left(x+3\right)}=0\)
<=> (3x - 1)(x + 3) - (2x + 5)(x - 1) - 8 = (x - 1)(x + 3)
<=> 3x2 + 9x - x - 3 - 2x2 + 2x - 5x + 5 - 8 = 0
<=> x2 + 5x - 6 = 0
<=> (x - 1)(x + 6) = 0
<=> x - 1 = 0 hoặc x + 6 = 0
<=> x = 1 (ktm) hoặc x = -6 (tm)
=> x = -6
\(\Leftrightarrow\frac{3x\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}-\frac{x\left(x-2\right)}{\left(x-2\right)\left(x-5\right)}+\frac{9x}{x^2-7x+10}=10\)
\(\Leftrightarrow\frac{3x^2-15x-x^2+2x+9x}{\left(x-2\right)\left(x-5\right)}=10\)
\(\Leftrightarrow2x^2-4x=10x^2-70x+100\)
\(\Leftrightarrow8x^2-66+100=0\)
\(\Leftrightarrow4x^2-33x+50=0\)
\(\Leftrightarrow4x\left(x-2\right)-25\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(4x-25\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=\frac{25}{4}\end{matrix}\right.\)
b) [(x-7)(x-2)][(x-4)(x-5)]=72
<=> (x2-9x+14)(x2-9x+20)=72
Đặt x2-9x+17=a
=> (a+3)(a-3)=72
<=> a2-9=72
<=> a2=81
=> a=\(\left\{9;-9\right\}\)
TH1: a=9
=> x2-9x+17=9
<=> x2-9x+8=0
<=> (x-1)(x-8)=0
=> x=\(\left\{1;8\right\}\)
TH2: a=-9
=> x2-9x+17=-9
<=> x2-9x+26=0
<=> x2-9x+20,25+5,75=0
<=> (x-4,5)2+5,75=0
=> x\(\in\varnothing\)
Vậy x=\(\left\{1;8\right\}\)
\(a.\frac{7x-3}{x-1}=\frac{2}{3}\\\Leftrightarrow \frac{3\left(7x-3\right)}{3\left(x-1\right)}= \frac{2\left(x-1\right)}{3\left(x-1\right)}\\ \Leftrightarrow3\left(7x-3\right)=2\left(x-1\right)\\\Leftrightarrow 3\left(7x-3\right)-2\left(x-1\right)=0\\ \Leftrightarrow21x-9-2x+2=0\\ \Leftrightarrow19x-7=0\\ \Leftrightarrow19x=7\\ \Leftrightarrow x=\frac{7}{19}\)
\(b.\frac{2\left(3-7x\right)}{1+x}=\frac{1}{2}\\ \Leftrightarrow\frac{4\left(3-7x\right)}{2\left(1+x\right)}=\frac{1\left(1+x\right)}{2\left(1+x\right)}\\\Leftrightarrow 4\left(3-7x\right)=1\left(1+x\right)\\ \Leftrightarrow4\left(3-7x\right)-1\left(1+x\right)=0\\ \Leftrightarrow12-28x-1-x=0\\ \Leftrightarrow11-29x=0\\ \Leftrightarrow-29x=-11\\ \Leftrightarrow x=\frac{-11}{-29}=\frac{11}{29}\)
\(c.\frac{5x-1}{3x+2}=\frac{5x-7}{3x-1}\\ \Leftrightarrow\frac{\left(5x-1\right)\left(3x-1\right)}{\left(3x+2\right)\left(3x-1\right)}=\frac{\left(5x-7\right)\left(3x+2\right)}{\left(3x+2\right)\left(3x-1\right)}\\ \Leftrightarrow\left(5x-1\right)\left(3x-1\right)=\left(5x-7\right)\left(3x+2\right)\\ \Leftrightarrow\left(5x-1\right)\left(3x-1\right)-\left(5x-7\right)\left(3x+2\right)=0\\ \Leftrightarrow15x^2-5x-3x+1-15x^2-10x+21x+14=0\\ \Leftrightarrow3x+15=0\\\Leftrightarrow 3x=-15\\\Leftrightarrow x=-5\)
\(d.\frac{4x+7}{x-1}=\frac{12x+5}{3x+4}\\\Leftrightarrow \frac{\left(4x+7\right)\left(3x+4\right)}{\left(x-1\right)\left(3x+4\right)}=\frac{\left(12x+5\right)\left(x-1\right)}{\left(3x+4\right)\left(x-1\right)}\\\Leftrightarrow \left(4x+7\right)\left(3x+4\right)=\left(12x+5\right)\left(x-1\right)\\\Leftrightarrow \left(4x+7\right)\left(3x+4\right)-\left(12x+5\right)\left(x-1\right)=0\\ \Leftrightarrow12x^2+16x+21x+28-12x^2-12x+5x-5=0\\ \Leftrightarrow30x+23=0\\ \Leftrightarrow30x=-23\\ \Leftrightarrow x=\frac{-23}{30}\)
\(e.\frac{1}{x-2}+3=\frac{3-x}{x-2}\\ \Leftrightarrow\frac{1}{x-2}+\frac{3\left(x-2\right)}{x-2}=\frac{3-x}{x-2}\\ \Leftrightarrow1+3\left(x-2\right)=3-x\\\Leftrightarrow 1+3x-6=3-x\\\Leftrightarrow 1+3x-6-3+x=0\\ \Leftrightarrow4x-8=0\\ \Leftrightarrow4x=8\\ \Leftrightarrow x=2\)
\(f.\frac{8-x}{x-7}-8=\frac{1}{x-7}\\ \Leftrightarrow\frac{8-x}{x-7}-\frac{8\left(x-7\right)}{x-7}=\frac{1}{x-7}\\ \Leftrightarrow8-x-8\left(x-7\right)=1\\ \Leftrightarrow8-x-8\left(x-7\right)-1=0\\\Leftrightarrow 8-x-8x+56-1=0\\\Leftrightarrow 63-9x=0\\\Leftrightarrow -9x=-63\\ \Leftrightarrow x=\frac{-63}{-9}=7\)
\(g.\frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{20}{x^2-25}\\ \Leftrightarrow\frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{20}{\left(x-5\right)\left(x+5\right)}\\\Leftrightarrow \frac{\left(x+5\right)\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\frac{\left(x-5\right)\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}=\frac{20}{\left(x-5\right)\left(x+5\right)}\\ \Leftrightarrow\left(x+5\right)\left(x+5\right)-\left(x-5\right)\left(x-5\right)=20\\\Leftrightarrow \left(x+5\right)\left(x+5\right)-\left(x-5\right)\left(x-5\right)-20=0\\ \Leftrightarrow x^2+5x+5x+25-x^2+5x+5x-25-20=0\\ \Leftrightarrow20x-20=0\\ \Leftrightarrow20x=20\\ \Leftrightarrow x=1\)
\(j.\frac{x}{2\left(x-3\right)}+\frac{x}{2\left(x+1\right)}=\frac{2x}{\left(x+1\right)\left(x-3\right)}\\\Leftrightarrow \frac{x\left(x+1\right)}{2\left(x+1\right)\left(x-3\right)}+\frac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}=\frac{2.2x}{2\left(x+1\right)\left(x-3\right)}\\ \Leftrightarrow x\left(x+1\right)+x\left(x-3\right)=4x\\\Leftrightarrow x\left(x+1\right)+x\left(x-3\right)-4x=0\\\Leftrightarrow x^2+x+x^2-3x-4x=0\\ \Leftrightarrow2x^2-6x=0\\ \Leftrightarrow2x\left(x-3\right)=0\\\Leftrightarrow\left[{}\begin{matrix}2x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right. \)
a) 8x - 3 = 5x + 12
<=> 8x - 5x = 12 + 3
<=> 3x = 15
<=> x = 5
b) \(\frac{x}{x^2-4}=\frac{1}{x+2}-\frac{1-x}{2-x}\) ; x khác +-2
<=> \(\frac{x}{\left(x-2\right)\left(x+2\right)}=\frac{1}{x+2}-\frac{1-x}{2-x}\)
=> x(2 - x) = (x - 2)(2 - x) - (1 - x)(x + 2)(x - 2)
<=> -x^2 + 2x = x^3 - 2x^2
<=> -x^2 + 2x - x^3 + 2x^2 = 0
<=> x^3 - x^2 - 2x = 0
<=> x(x + 1)(x - 2) = 0
<=> x = 0 hoặc x + 1 = 0 hoặc x - 2 = 0
<=> x = 0 (tm) hoặc x = -1 (tm) hoặc x = 2 (ktm)
Vậy: phương trình có tập nghiệm: S = {0; -1}
c) |x - 5| = 3x + 1
Ta có: \(\left|x-5\right|=\hept{\begin{cases}x-5\text{ nếu }x-5\ge0\Leftrightarrow x\ge5\\-\left(x-5\right)\text{ nếu }x-5< 0\Leftrightarrow x< 5\end{cases}}\)
+) Nếu x > 5, ta có phương trình:
x - 5 = 3x + 1
<=> x - 3x = 1 + 5
<=> -2x = 6
<=> x = -3 (ktm)
+) Nếu x < 5, ta có phương trình:
-(x - 5) = 3x + 1
<=> -x + 5 = 3x + 1
<=> -x - 3x = 1 - 5
<=> -4x = -4
<=> x = 1 (tm)
Vậy: phương trình có tập nghiệm: S = {1}
\(\frac{15x-10}{x^2+3}=0\)
\(\Leftrightarrow\frac{5\left(3x-2\right)}{x^2+3}=0\)
\(\Leftrightarrow5\left(3x-2\right)=0\)
\(\Leftrightarrow3x-2=0\)
\(\Leftrightarrow3x=2\)
\(\Leftrightarrow x=\frac{2}{3}\)
...
\(\frac{3x}{x-2}-\frac{x}{x-5}+\frac{9x}{x^2-7x+10}=10\)
\(\Rightarrow\frac{3x^2-15x-x^2+2x+9x}{x^2-7x+10}=10\)
\(\Rightarrow\frac{2x^2-4x}{x^2-7x+10}=10\)
\(\Rightarrow2x^2-4x=10x^2-70x+100\)
\(\Rightarrow8x^2-66x+100=0\)
Ta có \(\Delta=66^2-4.8.100=1156,\sqrt{\Delta}=34\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{66+34}{16}=\frac{25}{4}\\x=\frac{66-34}{16}=2\end{cases}}\)
a) \(\frac{3x}{x-2}-\frac{x}{x-5}+\frac{9x}{x^2-7x+10}=10\)
<=> \(\frac{3x\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}-\frac{x\left(x-2\right)}{\left(x-5\right)\left(x-2\right)}+\frac{9x}{\left(x-2\right)\left(x-5\right)}=10\)
<=> \(\frac{3x^2-15x-x^2+2x+9x}{\left(x-5\right)\left(x-2\right)}=10\)
<=> \(\frac{2x^2-4x}{\left(x-5\right)\left(x-2\right)}=10\)
<=> \(\frac{2x\left(x-2\right)}{\left(x-5\right)\left(x-2\right)}=10\)
<=> \(2x=10\left(x-5\right)\)
<=> 2x - 10x = -50
<=> -8x = -50
<=>x = 6,25
Vậy S = {6,25}
b) (x - 7)(x - 2)(x - 4)(x - 5) = 72
<=> (x2 - 9x + 14)(x2 - 9x + 20) = 72
Đặt x2 - 9x + 14 = t <=> t(t + 6) = 72
<=> t2 + 6t - 72 = 0
<=> t2 + 12t - 6t - 72 = 0
<=> (t + 12)(t - 6) = 0
<=> \(\orbr{\begin{cases}t+12=0\\t-6=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x^2-9x+14+12=0\\x^2-9x+14-6=0\end{cases}}\)
<=> \(\orbr{\begin{cases}\left(x-9x+20,25\right)+5,75=0\\x^2-9x+8=0\end{cases}}\)
<=> \(\orbr{\begin{cases}\left(x-4,5\right)^2+5,75=0\left(vn\right)\\x^2-x-8x+8=0\end{cases}}\)
<=> (x - 1)(x - 8) = 0
<=> \(\orbr{\begin{cases}x-1=0\\x-8=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=1\\x=8\end{cases}}\)
Vậy S = {1; 8}