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a) 7x - 35 = 0
<=> 7x = 0 + 35
<=> 7x = 35
<=> x = 5
b) 4x - x - 18 = 0
<=> 3x - 18 = 0
<=> 3x = 0 + 18
<=> 3x = 18
<=> x = 5
c) x - 6 = 8 - x
<=> x - 6 + x = 8
<=> 2x - 6 = 8
<=> 2x = 8 + 6
<=> 2x = 14
<=> x = 7
d) 48 - 5x = 39 - 2x
<=> 48 - 5x + 2x = 39
<=> 48 - 3x = 39
<=> -3x = 39 - 48
<=> -3x = -9
<=> x = 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
\(x-1=0\)
\(x=1\)
1) (2x - 3)2 = 4x2 - 8
<=> 4x2 - 12x + 9 = 4x2 - 8
<=> 12x + 9 = -8
<=> 12x = -17
<=> x = 17/12
1) (2x - 3)^2 = 4x^2 - 8
<=> 4x^2 - 12x + 9 = 4x^2 - 8
<=> 4x^2 - 12x + 9 - 4x^2 = -8
<=> -12x + 9 = -8
<=> -12x = -8 - 9
<=> -12x = -17
<=> x = 17/12
2) x - (x + 2)(x - 3) = 4 - x^2
<=> x - x^2 + 3x - 2x + 6 = 4 - x^2
<=> 2x - x^2 + 6 = 4 - x^2
<=> 2x - x^2 + 6 + x^2 = 4
<=> 2x + 6 = 4
<=> 2x = 4 + 6
<=> 2x = 10
<=> x = 5
3) 3x - (x - 3)(x + 1) = 6x - x^2
<=> 3x - x^2 - x + 3x + 3 = 6x - x^2
<=> 5x - x^2 + 3 = 6x - x^2
<=> 5x - x^2 + 3 + x^2 = 6x
<=> 5x + 3 = 6x
<=> 3 = 6x - 5x
<=> 3 = x
4) 3x/4 = 6
<=> 3x = 6.4
<=> 3x = 24
<=> x = 8
5) 7 + 5x/3 = x - 2
<=> 21 + 5x = 3x - 6
<=> 5x = 3x - 6 - 21
<=> 5x = 3x - 27
<=> 5x - 3x = -27
<=> 2x = -27
<=> x = -27/2
6) x + 4 = 2/5x - 3
<=> 5x + 20 = 2x - 15
<=> 5x + 20 - 2x = -15
<=> 3x + 20 = -15
<=> 3x = -15 - 20
<=> 3x = -35
<=> x = -35/3
7) 1 + x/9 = 4/3
<=> x/9 = 4/3 - 1
<=> x/9 = 1/3
<=> x = 3
\(a,\Leftrightarrow5\left(x-2\right)-15x\le9+10\left(x+1\right)\)
\(\Leftrightarrow5x-10-15x\le9+10x+10\)
\(\Leftrightarrow-20x\le29\)
\(\Leftrightarrow x\ge-1,45\)
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 ..............
\(\frac{x-2}{6}-\frac{x}{2}\le\frac{3}{10}+\frac{x+1}{3}\Leftrightarrow\frac{5\left(x-2\right)}{30}-\frac{15x}{30}\le\frac{9}{30}+\frac{10\left(x+1\right)}{30}\)
\(\Leftrightarrow5x-10-15x-9-10x-10\le0\)
\(\Leftrightarrow-20x-29\le0\Leftrightarrow\left(-20x\right)\cdot\frac{-1}{20}\ge29\cdot-\frac{1}{20}\)
\(\Leftrightarrow x\ge-\frac{29}{20}\)
Nhìn sơ qua thì thấy bài 3, b thay -2 vào x rồi giải bình thường tìm m
Bài 2:
a) \(x+x^2=0\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x=0\\x+1=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=0\\x=0-1\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=0\\x=-1\end{cases}}\)
b) \(0x-3=0\)
\(\Leftrightarrow0x=3\)
\(\Rightarrow vonghiem\)
c) \(3y=0\)
\(\Leftrightarrow y=0\)
a, 8/x-8 + 11/x-11 = 9/x-9 + 10/ x-10
b, x/x-3 - x/x-5 = x/x-4 - x/x-6
c, 4/x^2-3x+2 - 3/2x^2-6x+1 +1 = 0
d, 1/x-1 + 2/ x-2 + 3/x-3 = 6/x-6
e, 2/2x+1 - 3/2x-1 = 4/4x^2-1
f, 2x/x+1 + 18/x^2+2x-3 = 2x-5 /x+3
g, 1/x-1 + 2x^2 -5/x^3 -1 = 4/ x^2 +x+1
\(\frac{x^2+5}{25-x^2}=\frac{3}{x+5}+\frac{x}{x-5}\)
\(\Leftrightarrow\frac{x^2+5}{\left(5-x\right)\left(5+x\right)}=\frac{3}{5+x}-\frac{x}{5-x}\)
\(\Leftrightarrow\frac{x^2+5}{\left(5-x\right)\left(5+x\right)}=\frac{3\left(5-x\right)-x\left(5+x\right)}{\left(5-x\right)\left(5+x\right)}\)
\(\Rightarrow x^2+5=3\left(5-x\right)-x\left(5+x\right)\)
\(\Leftrightarrow x^2+5=15-3x-5x-x^2\)
\(\Leftrightarrow15-3x-5x-x^2-x^2-5=0\)
\(\Leftrightarrow10-8x-2x^2=0\)
\(\Leftrightarrow2x^2+8x-10=0\)
\(\Leftrightarrow2\left(x^2+4x-5\right)=0\)
\(\Leftrightarrow2\left(x^2+5x-x-5\right)=0\)
\(\Leftrightarrow x^2-x+5x-5=0\)
\(\Leftrightarrow x\left(x-1\right)+5\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-5\end{cases}}}\)
