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Bài 1:
1. \(x-8=3-2\left(x+4\right)\)
\(x-8=3-2x-8\)
\(3x=3\Rightarrow x=1\)
2. \(2\left(x+3\right)-3\left(x-1\right)=2\)
\(2x+6-3x+3=2\)
\(-x+9=2\Rightarrow x=7\)
3. \(4\left(x-5\right)-\left(3x-1\right)=x-19\)
\(4x-20-3x+1=x-19\)
\(0x=0\Rightarrow x=0\)
4. \(7-\left(x-2\right)=5\left(2x-3\right)\)
\(7-x+2=10x-15\)
\(-11x=-24\Rightarrow x=\frac{24}{11}\)
5. \(32-4\left(0,5y-5\right)=3y+2\)
\(32-2y+20=3y+2\)
\(-5y=-50\Rightarrow y=10\)
6. \(3\left(x-1\right)-x=2x-3\)
\(3x-3-x=2x-3\)
\(0x=0\Rightarrow x=0\)
Bài 2:
1. \(\frac{2-x}{3}=\frac{3-2x}{5}\)
\(\frac{\left(2-x\right)5}{15}-\frac{\left(3-2x\right)3}{15}=0\)
\(\frac{10-5x-9+6x}{15}=0\)
\(x+1=0\Rightarrow x=-1\)
2. \(\frac{3-4x}{4}=\frac{x+2}{5}\)
\(\frac{5\left(3-4x\right)}{20}-\frac{4\left(x+2\right)}{20}=0\)
\(\frac{15-20x-4x-8}{20}=0\)
\(7-24x=0\)
\(24x=7\Rightarrow x=\frac{7}{24}\)
Mình ko ghi lại đề , bạn ghi ra xong rồi suy ra như mình nha .
1) \(=>A=\left(6x^2+3x-10x-5\right)-\left(6x^2+14x-9x-21\right)\)
\(=>A=-12x+16\)
2) \(=>B=8x^3+27-8x^3+2=29\)
3)\(=>C=[\left(x-1\right)-\left(x+1\right)]^3=\left(-2\right)^3=-8\)
4)\(=>D=[\left(2x+5\right)-\left(2x\right)]^3=5^3=125\)
5)\(=>E=\left(3x+1\right)^2-\left(3x+5\right)^2+12x+2\left(6x+3\right)\)
\(=>E=\left(3x+1+3x+5\right)\left(3x+1-3x-5\right)+12x+12x+6\)
\(=>E=\left(6x+6\right)\left(-4\right)+24x+6=-24x-24+24x+6=-18\)
6)\(=>F=\left(2x^2+3x-10x-15\right)-\left(2x^2-6x\right)+x+7=-8\)
k cho mik nha ,
1: Ta có: \(2x\left(x+3\right)-6\left(x-3\right)=0\)
\(\Leftrightarrow2x^2+6x-6x+18=0\)
\(\Leftrightarrow2x^2+18=0\left(loại\right)\)
2: Ta có: \(2x^2\left(2x+3\right)+\left(2x+3\right)=0\)
\(\Leftrightarrow2x+3=0\)
hay \(x=-\dfrac{3}{2}\)
3: Ta có: \(\left(x-2\right)\left(x+1\right)-4x\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(1-3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)
4: Ta có: \(2x\left(x-5\right)-3x+15=0\)
\(\Leftrightarrow\left(x-5\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{3}{2}\end{matrix}\right.\)
5: Ta có: \(3x\left(x+4\right)-2x-8=0\)
\(\Leftrightarrow\left(x+4\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{2}{3}\end{matrix}\right.\)
6: Ta có: \(x^2\left(2x-6\right)+2x-6=0\)
\(\Leftrightarrow2x-6=0\)
hay x=3
\(a.\left(2x+5\right)\frac{6}{2}=75\\ \Leftrightarrow\left(2x+5\right)3=75\\ \Leftrightarrow6x+15=75\\\Leftrightarrow 6x=75-15\\\Leftrightarrow 6x=60\\ \Leftrightarrow x=10\)
\(b.\frac{x-3}{5}=6-\frac{1-2x}{3}\\ \Leftrightarrow\frac{3\left(x-3\right)}{15}=\frac{6.15}{15}-\frac{5\left(1-2x\right)}{15}\\ \Leftrightarrow3\left(x-3\right)=6.15-5\left(1-2x\right)\\ \Leftrightarrow3x-9=90-5+10x\\ \Leftrightarrow3x-9-90+5-10x=0\\ \Leftrightarrow-7x-94=0\\ \Leftrightarrow-7x=94\\ \Leftrightarrow x=\frac{-94}{7}\)
\(c.\frac{2x}{3}+\frac{2x-1}{6}=\frac{4-x}{3}\\ \Leftrightarrow\frac{2x.2}{6}+\frac{2x-1}{6}=\frac{2\left(4-x\right)}{6}\\ \Leftrightarrow2x.2+2x-1=2\left(4-x\right)\\ \Leftrightarrow4x+2x-1=8-2x\\ \Leftrightarrow4x+2x-1-8+2x=0\\ \Leftrightarrow8x-9=0\\ \Leftrightarrow8x=9\\ \Leftrightarrow x=\frac{9}{8}\)
\(d.\frac{x-1}{2}+\frac{x-1}{4}=\frac{1-x}{3}\\ \Leftrightarrow\frac{6\left(x-1\right)}{12}+\frac{3\left(x-1\right)}{12}=\frac{4\left(1-x\right)}{12}\\ \Leftrightarrow6\left(x-1\right)+3\left(x-1\right)=4\left(1-x\right)\\ \Leftrightarrow6x-6+3x-3=4-4x\\ \Leftrightarrow6x-6+3x-3-4+4x=0\\ \Leftrightarrow13x-13=0\\ \Leftrightarrow13x=13\\ \Leftrightarrow x=1\)
b: =>1/4x+4/5-x-5=1/3x+1-1/2x+1
=>-3/4x+1/6x=2+5-4/5=24/5
=>x=-288/35
c: =>6x^2+3x-30x-15=6x^2+10x-21x-35
=>-27x-15=-11x-35
=>-16x=-20
=>x=5/4
\(1,\dfrac{4x-4}{3}=\dfrac{7-x}{5}\\ \Leftrightarrow5\left(4x-4\right)=3\left(7-x\right)\\ \Leftrightarrow20x-20=21-3x\\ \Leftrightarrow17x=41\Leftrightarrow x=\dfrac{41}{17}\)
\(2,\dfrac{3x-9}{5}=\dfrac{3-x}{2}\\ \Leftrightarrow6x-18=15-5x\\ \Leftrightarrow11x=33\\ \Leftrightarrow x=3\)
\(3,\dfrac{2x-1}{5}-\dfrac{3-x}{3}=1\\ \Leftrightarrow\dfrac{6x-3-15+5x}{15}=1\\ \Leftrightarrow11x-18=1\\ \Leftrightarrow x=\dfrac{19}{11}\)
\(4,\dfrac{x-5}{3}+\dfrac{3x+4}{2}=\dfrac{5x+2}{6}\\ \Leftrightarrow2x-10+9x+12=5x+2\\ \Leftrightarrow6x=0\Leftrightarrow x=0\)
\(5,\dfrac{x-3}{2}+\dfrac{2x+3}{5}=\dfrac{2x+5}{10}\\ \Leftrightarrow5x-15+4x+6=2x+5\\ \Leftrightarrow7x=14\\ \Leftrightarrow x=2\)
Tick nha
2: Ta có: \(\dfrac{3x-9}{5}=\dfrac{3-x}{2}\)
\(\Leftrightarrow6x-18=15-5x\)
\(\Leftrightarrow11x=33\)
hay x=3
\(a,\dfrac{3}{2x-1}+1=\dfrac{2x-1}{2x+1};ĐKXĐ:x\ne\pm\dfrac{1}{2}\\ \Leftrightarrow\dfrac{3}{2x-1}-\dfrac{2x-1}{2x+1}+1=0\\ \Leftrightarrow\dfrac{3\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)}-\dfrac{\left(2x-1\right)\left(2x-1\right)}{\left(2x+1\right)\left(2x-1\right)}+\dfrac{\left(2x-1\right)\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)}=0\\ \Rightarrow3\left(2x+1\right)-\left(2x-1\right)^2+\left(2x-1\right)\left(2x+1\right)=0\\ \Leftrightarrow6x+3-\left(4x^2-4x+1\right)+\left(4x^2-1\right)=0\\ \Leftrightarrow6x+3-4x^2+4x-1+4x^2-1=0\\ \Leftrightarrow10x+1=0\\ \Leftrightarrow10x=-1\\ \Leftrightarrow x=-\dfrac{1}{10}\)
Vậy \(x\in\left\{-\dfrac{1}{10}\right\}\)
2:
a: =>-2x=10
=>x=-5
b: =>(x-3)(2x+5)=0
=>x=3 hoặc x=-5/2
Th1 là trường hợp 1 ,Th2 là trường hợp 2 chứ ko phải là cách 1 cách 2 đâu
Chắc chắn là có thể.
Em lớp 6, học dạng này quen rồi