Tìm số nguyên x biết
1,( 4x+1)/(3x-2)=-5/4
2, (x+5)/-7=(2x+3)/4
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Tìm x biết
1. 2(5x-8)-3(4x-5)=4(3x-4)+11
2. (2x+1)2-(4x-1).(x-3)-15=0
3. (3x-1).(2x-7)-(1-3x).(6x-5)=0
1) \(\Rightarrow10x-16-12x+15=12x-16+11\)
\(\Rightarrow14x=4\Rightarrow x=\dfrac{2}{7}\)
2) \(\Rightarrow4x^2+4x+1-4x^2+13x-3-15=0\)
\(\Rightarrow17x=17\Rightarrow x=1\)
3) \(\Rightarrow\left(3x-1\right)\left(2x-7+6x-5\right)=0\)
\(\Rightarrow\left(2x-3\right)\left(3x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
2: Ta có: \(\left(2x+1\right)^2-\left(4x-1\right)\left(x-3\right)-15=0\)
\(\Leftrightarrow4x^2+4x+1-4x^2+12x+x-3-15=0\)
\(\Leftrightarrow17x=17\)
hay x=1
Noob ơi, bạn phải đưa vào máy tính ý solve cái là ra x luôn, chỉ tội là đợi hơi lâu
a, 4.(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x + 14)
=> 72 - 20x - 36x + 84 = 30x - 240 - 6x - 84
=> (72 + 84) + (-20x - 36x) = (30x - 6x) + (-240 - 84)
=> 156 - 56x = 24x - 324
=> 24x + 56x = 324 + 156
=> 80x = 480
=> x = 480 : 80 = 6
Vậy x = 6
a) \(3^{x+2}\cdot5^{y-3}=45^x\)
\(\Rightarrow3^{x+2}\cdot5^{y-3}=\left(3^2\right)^x\cdot5^x\)
\(\Rightarrow3^{x+2}\cdot5^{y-3}=3^{2x}\cdot5^x\)
\(\Rightarrow\left\{{}\begin{matrix}3^{x+2}=3^{2x}\\5^{y-3}=5^x\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+2=2x\\y-3=x\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=2\\y-3=2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=2\\y=5\end{matrix}\right.\)
1: \(\frac{3x-2}{3}-2=\frac{4x+1}{4}\)
=>\(\frac{3x-2-6}{3}=\frac{4x+1}{4}\)
=>\(\frac{3x-8}{3}=\frac{4x+1}{4}\)
=>3(4x+1)=4(3x-8)
=>12x+3=12x-32
=>3=-32(vô lý)
=>Phương trình vô nghiệm
2: \(\frac{x-3}{4}+\frac{2x-1}{3}=\frac{2-x}{6}\)
=>\(\frac{3\left(x-3\right)+4\left(2x-1\right)}{12}=\frac{2\left(2-x\right)}{12}\)
=>3(x-3)+4(2x-1)=2(2-x)
=>3x-9+8x-4=4-2x
=>11x-13=4-2x
=>13x=17
=>\(x=\frac{17}{13}\)
3: \(\frac12\left(x+1\right)+\frac14\left(x+3\right)=3-\frac13\left(x+2\right)\)
=>\(\frac12x+\frac12+\frac14x+\frac34+\frac13x+\frac23=3\)
=>\(x\left(\frac12+\frac14+\frac13\right)+\frac{6}{12}+\frac{9}{12}+\frac{8}{12}=3\)
=>\(x\left(\frac{6}{12}+\frac{3}{12}+\frac{4}{12}\right)=3-\frac{23}{12}=\frac{36}{12}-\frac{23}{12}=\frac{13}{12}\)
=>\(x\cdot\frac{13}{12}=\frac{13}{12}\)
=>x=1
4: \(\frac{x+4}{5}-x+4=\frac{x}{3}-\frac{x-2}{2}\)
=>\(\frac{x+4}{5}+\frac{5\left(-x+4\right)}{5}=\frac{2x-3\left(x-2\right)}{6}\)
=>\(\frac{x+4-5x+20}{5}=\frac{2x-3x+6}{6}\)
=>\(\frac{-4x+24}{5}=\frac{-x+6}{6}\)
=>6(-4x+24)=5(-x+6)
=>-24x+144=-5x+30
=>-19x=-114
=>x=6
5: \(\frac{4-5x}{6}=\frac{2\left(-x+1\right)}{2}\)
=>\(\frac{4-5x}{6}=-x+1\)
=>6(-x+1)=-5x+4
=>-6x+6=-5x+4
=>-6x+5x=4-6
=>-x=-2
=>x=2
6: \(-\left(\frac{x-3}{2}-2\right)=\frac{5\left(x+2\right)}{4}\)
=>\(-\frac{x-3-4}{2}=\frac{5\left(x+2\right)}{4}\)
=>\(\frac{-2\left(x-7\right)}{4}=\frac{5\left(x+2\right)}{4}\)
=>5(x+2)=-2(x-7)
=>5x+10=-2x+14
=>7x=4
=>x=4/7
7: \(\frac{2\left(2x+1\right)}{5}-\frac{6+x}{3}=\frac{5-4x}{15}\)
=>\(\frac{6\left(2x+1\right)-5\left(x+6\right)}{15}=\frac{5-4x}{15}\)
=>6(2x+1)-5(x+6)=-4x+5
=>12x+6-5x-30=-4x+5
=>7x-24=-4x+5
=>7x+4x=5+24
=>11x=29
=>\(x=\frac{29}{11}\)
8: \(\frac{7-3x}{2}-\frac{5+x}{5}=1\)
=>\(\frac{5\left(7-3x\right)-2\left(x+5\right)}{10}=1\)
=>5(7-3x)-2(x+5)=10
=>35-15x-2x-10=10
=>-17x+25=10
=>-17x=-15
=>x=15/17
Rút gọn hết ta được :
a/ 41x - 17 = -21
=> 41x = -4 => x = 4/41
b/ 34x - 17 = 0
=> 34x = 17
=> x = 17/34 = 1/2
c/ 19x + 56 = 52
=> 19x = -4
=> x = -4/19
d/ 20x2 - 16x - 34 = 10x2 + 3x - 34
=> 10x2 - 19x = 0
=> x(10x - 19) = 0
=> x = 0
hoặc 10x - 19 = 0 => 10x = 19 => x = 19/10
Vậy x = 0 ; x = 19/10
Rút gọn hết ta được :
a/ 41x - 17 = -21
=> 41x = -4 => x = 4/41
b/ 34x - 17 = 0
=> 34x = 17
=> x = 17/34 = 1/2
c/ 19x + 56 = 52
=> 19x = -4
=> x = -4/19
d/ 20x 2 - 16x - 34 = 10x 2 + 3x - 34
=> 10x 2 - 19x = 0
=> x(10x - 19) = 0
=> x = 0 hoặc 10x - 19 = 0
=> 10x = 19
=> x = 19/10
Vậy x = 0 ; x = 19/10
\(1.\frac{4x+1}{3x-2}=-\frac{5}{4}\)
\(\Rightarrow\left(4x+1\right)\times4=\left(-5\right)\times\left(3x-2\right)\)
\(16x+4=\left(-15x\right)+10\)
\(16x+15x=10-4\)
\(31x=6\)
\(x=\frac{6}{31}\)
Vậy \(x\in\left\{\frac{6}{31}\right\}\)
\(2.\frac{x+5}{-7}=\frac{2x+3}{4}\)
\(\Rightarrow\left(x+5\right)\times4=\left(2x+3\right)\times\left(-7\right)\)
\(4x+20=\left(-14x\right)+\left(-21\right)\)
\(4x+14x=\left(-21\right)-20\)
\(18x=-41\)
\(x=-\frac{41}{18}\)
Vậy \(x\in\left\{-\frac{41}{18}\right\}\)