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a)\(60x^2+35x-60x^2+15x=100\)
35x+15x=100
50x=100 =>x=2
b)\(10x^2-35x+16x-10x^2=5\)
-35x+16x=5
-19x=5 =>x=-5/19
Câu 1:
\(\frac{2x-1}{3}=\frac{3x+1}{4}\)
\(\frac{8x-4}{12}=\frac{9x+3}{12}\)
\(\Rightarrow8x-4=9x+3\)
\(8x-9x=3+4\)
\(-x=7\Rightarrow x=-7\)
Câu 2:
\(\frac{3x}{5x+4}=\frac{12}{28}\)
\(\Rightarrow\left(3x\right).28=12.\left(5x+4\right)\)
\(84x=60x+48\)
\(84x-60x=48\)
\(24x=48\)
\(\Leftrightarrow x=2\)
Câu 3 :
\(\frac{x}{2}=\frac{8}{x}\)
\(\Rightarrow x.x=8.2\)
\(x^2=16\)
\(x^2=\left(\pm4\right)^2\)
\(\Rightarrow x\left\{-4;4\right\}\)
học tốt !!!
Câu 1: \(\frac{2x-1}{3}=\frac{3x+1}{4}\)
\(\Leftrightarrow2\left(2x-1\right)=3\left(3x+1\right)\)
\(\Leftrightarrow4x-2=9x+3\)
<=> -9x+4x=3+2
<=> -5x=5
<=> x=-1
Câu 3: \(\frac{x}{2}=\frac{8}{x}\left(x\ne0\right)\)
<=> x2=16
<=> \(\orbr{\begin{cases}x=4\\x=-4\end{cases}\left(tmđk\right)}\)
Câu 2: Bạn cũng nhân chéo lên nhưng nhớ đi điều kiện: \(x\ne\frac{-4}{5}\)
\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)
\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)
Bài 4:
a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)
\(\Leftrightarrow6x-9-2x+4=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
\(\Leftrightarrow3x=13\)
hay \(x=\dfrac{13}{3}\)
c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
\(\Leftrightarrow-8x=-8\)
hay x=1
\(5x-\frac{1}{3}=3x+\frac{2}{7}=8-\frac{5x}{2}\)
\(\Leftrightarrow5x-3x=\frac{2}{7}+\frac{1}{3}\)
\(\Leftrightarrow2x=\frac{13}{21}\)
\(\Leftrightarrow x=\frac{13}{42}\)
Thử lại:
\(5x-\frac{1}{3}=5\cdot\frac{13}{42}-\frac{1}{3}=\frac{17}{14}\)
\(3x+\frac{2}{7}=3\cdot\frac{13}{42}+\frac{2}{7}=\frac{17}{14}\)
\(8-\frac{5x}{2}=8-5\cdot\frac{13}{42}\div2=\frac{607}{84}\)( vô lý)
Vậy không có giá trị nào của x thoả mãn
a)x+15=-7
x=(-7)-15
x=-22
b)\(\left(5x-2\right).8^3=8^4\)
(5x-2).512=4096
(5x-2)=4096:512
(5x-2)=8
5x=8+2
5x=10
x=10:5
x=2
c)|x-1|=2
=>x-1=2 hoac x-1=-2
x=2+1 x=-2+1
x=3 x=-1
a/ => x = (-7) - 15 = -22
Vậy x = -22
b/ (5x - 2) . 83 = 84
=> 5x - 2 = 8
=> 5x = 10
=> x = 2
Vậy x = 2
c/ => x - 1 = 2 => x = 3
hoặc x - 1 = -2 => x = -1
Vậy x = 3 ; x = -1
a)4(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x + 14)
<=>72 - 20x - 36x +84 = 30x - 240 - 6x 84
<=> -80x = -480
<=> x = 6
b) 5(3x+5)-4(2x-3) =5x+3(2x+12)+1
<=> 15x + 25 - 8x + 12 = 5x + 6x + 36 + 1
<=> 15x + 25 - 8x + 12 - 5x - 6x - 36 - 1 = 0
<=> -4x = 0
<=> x = 0
c) 2(5x-8)-3(4x-5)=4(3x-4)+11
= 10x - 16 - 12x + 15 = 12x - 16 + 11
= -14x = -4
= x =\(\frac{2}{7}\)
d) 5x-3{4x-2[4x-3(5x-2)]}=182
= 5x - 3 . [4x - 2(4x - 15x + 6)]
= 5x - 3 . (4x - 8x + 30x - 12)
= 5x - 12x + 24x - 90x + 36
= -73x + 36 = 182
=> -73x = 182 - 36 = 146
=> x = 146 : (-73) = -2
~Hok tốt~
5\(^{x+1}\) - 5\(^x\) = 2.28 + 8
5\(^x\).(5 - 1) = 520
5\(^x\).4 = 520
5\(^x\) = 520 : 4
5\(^x\) = 130
Với \(x\) = 0 ⇒ 5\(^x\) = 50 = 1 < 130 (loại)
Với \(x\) > 0 ⇒ 5\(^x\) = \(\overline{...5}\) \(\ne\) 130 (loại)
Vậy \(x\) \(\in\) \(\varnothing\)
\(5^{x+1}-5^x=2.2^8+8\\ 5^x\left(5-1\right)=512+8\\ 5^x.4=520\\ 5^x=\dfrac{520}{4}=130\)
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