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KH
22 tháng 9 2018
* Trả lời:
\(\left(1\right)\) \(-3\left(1-2x\right)-4\left(1+3x\right)=-5x+5\)
\(\Leftrightarrow-3+6x-4-12x=-5x+5\)
\(\Leftrightarrow6x-12x+5x=3+4+5\)
\(\Leftrightarrow x=12\)
\(\left(2\right)\) \(3\left(2x-5\right)-6\left(1-4x\right)=-3x+7\)
\(\Leftrightarrow6x-15-6+24x=-3x+7\)
\(\Leftrightarrow6x+24x+3x=15+6+7\)
\(\Leftrightarrow33x=28\)
\(\Leftrightarrow x=\dfrac{28}{33}\)
\(\left(3\right)\) \(\left(1-3x\right)-2\left(3x-6\right)=-4x-5\)
\(\Leftrightarrow1-3x-6x+12=-4x-5\)
\(\Leftrightarrow-3x-6x+4x=-1-12-5\)
\(\Leftrightarrow-5x=-18\)
\(\Leftrightarrow x=\dfrac{18}{5}\)
\(\left(4\right)\) \(x\left(4x-3\right)-2x\left(2x-1\right)=5x-7\)
\(\Leftrightarrow4x^2-3x-4x^2+2x=5x-7\)
\(\Leftrightarrow-x-5x=-7\)
\(\Leftrightarrow-6x=-7\)
\(\Leftrightarrow x=\dfrac{7}{6}\)
\(\left(5\right)\) \(3x\left(2x-1\right)-6x\left(x+2\right)=-3x+4\)
\(\Leftrightarrow6x^2-3x-6x^2-12x=-3x+4\)
\(\Leftrightarrow-15x+3x=4\)
\(\Leftrightarrow-12x=4\)
\(\Leftrightarrow x=-\dfrac{1}{3}\)
18 tháng 6 2019
a) \(\left|x+9\right|=2x\)
\(\Leftrightarrow\left[{}\begin{matrix}x+9=2x\\x+9=-2x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-3\end{matrix}\right.\)
b) \(\left|5x\right|-3x=2\Leftrightarrow\left|5x\right|=3x+2\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=3x+2\\-5x=3x+2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{-1}{4}\end{matrix}\right.\)
c) \(\left|x+6\right|-9=2x\Leftrightarrow\left|x+6\right|=2x+9\)
\(\Leftrightarrow\left[{}\begin{matrix}x+6=2x+9\\-x-6=2x+9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-5\end{matrix}\right.\)
d) \(\left|2x-3\right|+x=21\Leftrightarrow\left|2x-3\right|=21-x\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=21-x\\2x-3=x-21\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-18\end{matrix}\right.\)
e) \(\left|2x+4\right|=-4x\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+4=4x\\2x+4=-4x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{-2}{3}\end{matrix}\right.\)
i) \(\left|3x-1\right|+2=x\Leftrightarrow\left|3x-1\right|=x-2\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=x-2\\3x-1=2-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{2}\\x=\frac{3}{4}\end{matrix}\right.\)
g) \(\left|x+15\right|+1=3x\Leftrightarrow\left|x+15\right|=3x-1\)
\(\Leftrightarrow\left[{}\begin{matrix}x+15=3x-1\\x+15=1-3x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-3,5\end{matrix}\right.\)
h) \(\left|2x-5\right|+x=2\Leftrightarrow\left|2x-5\right|=2-x\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=2-x\\2x-5=x-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{3}\\x=3\end{matrix}\right.\)
QD
18 tháng 6 2019
a) |9+x|=2x
TH1: 9+x=2x
<=> 9=2x-x
<=> x=9
TH2: -9-x=2x
<=> -9=3x
<=> x=-3
b) |5x|-3x=2
TH1: 5x-3x=2
<=> 2x=2
<=> x=1
TH2: -5x-3x=2
<=> -8x=2
<=>x=-4
c) |x+6|-9=2x
TH1: x+6-9=2x
<=> -3=x
TH2: -x-6-9=2x
<=> -15=3x
<=>x=-5
d) |2x-3|+x=21
TH1: 2x-3+x=21
<=> 3x=24
<=> x=8
TH2: -2x+3+x=21
<=> -x=18
<=> x=-18
e,i,g,h tương tự
25 tháng 8 2020
a) \(2x+\frac{3}{15}=\frac{7}{5}\)
=> \(2x=\frac{7}{5}-\frac{3}{15}=\frac{21}{15}-\frac{3}{15}=\frac{18}{15}\)
=> \(x=\frac{18}{15}:2=\frac{18}{15}\cdot\frac{1}{2}=\frac{9}{15}\cdot\frac{1}{1}=\frac{9}{15}\)
b) \(x-\frac{2}{9}=\frac{8}{3}\)
=> \(x=\frac{8}{3}+\frac{2}{9}\)
=> \(x=\frac{24}{9}+\frac{2}{9}=\frac{26}{9}\)
c) \(\frac{-8}{x}=\frac{-x}{18}\)
=> x(-x) = (-8).18
=> -x2 = -144
=> x2 = 144(bỏ dấu âm)
=> x = \(\pm\)12
d) \(\frac{2x+3}{6}=\frac{x-2}{5}\)
=> 5(2x + 3) = 6(x - 2)
=> 10x + 15 = 6x - 12
=> 10x + 15 - 6x + 12 = 0
=> 4x + 27 = 0
=> 4x = -27
=> x = -27/4
e) \(\frac{x+1}{22}=\frac{6}{x}\)
=> x(x + 1) = 132
=> x(x + 1) = 11.12
=> x = 11
f) \(\frac{2x-1}{2}=\frac{5}{x}\)
=> x(2x - 1) = 10
=> 2x2 - x = 10
=> 2x2 - x - 10 = 0
tới đây tự làm đi nhé
g) \(\frac{2x-1}{21}=\frac{3}{2x+1}\)
=> (2x - 1)(2x + 1) = 63
=> 4x2 - 1 = 63
=> 4x2 = 64
=> x2 = 16
=> x = \(\pm\)4
h) Tương tự
NC
25 tháng 8 2020
a) \(\frac{2x+3}{15}=\frac{7}{5}\Leftrightarrow10x+15=105\Leftrightarrow10x=90\Rightarrow x=9\)
b) \(\frac{x-2}{9}=\frac{8}{3}\Leftrightarrow3x-6=72\Leftrightarrow3x=78\Rightarrow x=26\)
c) \(\frac{-8}{x}=\frac{-x}{18}\Leftrightarrow x^2=144\Leftrightarrow\orbr{\begin{cases}x=12\\x=-12\end{cases}}\)
d) \(\frac{2x+3}{6}=\frac{x-2}{5}\Leftrightarrow10x+15=12x-12\Leftrightarrow2x=27\Rightarrow x=\frac{27}{2}\)
e) \(\frac{x+1}{22}=\frac{6}{x}\Leftrightarrow x^2+x-132=0\Leftrightarrow\left(x-11\right)\left(x+12\right)=0\Leftrightarrow\orbr{\begin{cases}x=11\\x=-12\end{cases}}\)
f) \(\frac{2x-1}{2}=\frac{5}{x}\Leftrightarrow2x^2-x-10=0\Leftrightarrow\left(x-2\right)\left(2x+5\right)=0\Leftrightarrow\orbr{\begin{cases}x=2\\x=-\frac{5}{2}\end{cases}}\)
g) \(\frac{2x-1}{21}=\frac{3}{2x+1}\Leftrightarrow4x^2=64\Leftrightarrow x^2=16\Rightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
h) \(\frac{10x+5}{6}=\frac{5}{x+1}\Leftrightarrow10x^2+15x-25=0\Leftrightarrow5\left(x-1\right)\left(2x+5\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{5}{2}\end{cases}}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
27 tháng 9 2024
Bài 1:
- \(\dfrac{11}{2}x\) + 1 = \(\dfrac{1}{3}x-\dfrac{1}{4}\)
- \(\dfrac{11}{2}\)\(x\) - \(\dfrac{1}{3}\)\(x\) = - \(\dfrac{1}{4}\) - 1
-(\(\dfrac{33}{6}\) + \(\dfrac{2}{6}\))\(x\) = - \(\dfrac{5}{4}\)
- \(\dfrac{35}{6}\)\(x\) = - \(\dfrac{5}{4}\)
\(x=-\dfrac{5}{4}\) : (- \(\dfrac{35}{6}\))
\(x\) = \(\dfrac{3}{14}\)
Vậy \(x=\dfrac{3}{14}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
27 tháng 9 2024
Bài 2: 2\(x\) - \(\dfrac{2}{3}\) - 7\(x\) = \(\dfrac{3}{2}\) - 1
2\(x\) - 7\(x\) = \(\dfrac{3}{2}\) - 1 + \(\dfrac{2}{3}\)
- 5\(x\) = \(\dfrac{9}{6}\) - \(\dfrac{6}{6}\) + \(\dfrac{4}{6}\)
- 5\(x\) = \(\dfrac{7}{6}\)
\(x\) = \(\dfrac{7}{6}\) : (- 5)
\(x\) = - \(\dfrac{7}{30}\)
Vậy \(x=-\dfrac{7}{30}\)
PN
11 tháng 2 2018
a) \(\left|9+x\right|=2x\)
\(\Rightarrow\left[{}\begin{matrix}9+x=2x\\9+x=-2x\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x-x=9\\-2x-x=9\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=9\\-3x=9\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=9\\x=-3\end{matrix}\right.\)
b) \(\left|5x\right|-3x=2\)
\(\Leftrightarrow\left|5x\right|=2+3x\)
\(\Rightarrow\left[{}\begin{matrix}5x=2+3x\\5x=-2-3x\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}5x-3x=2\\5x+3x=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=2\\8x=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{4}\end{matrix}\right.\)
c) \(\left|x+6\right|-9=2x\)
\(\Leftrightarrow\left|x+6\right|=2x+9\)
\(\Rightarrow\left[{}\begin{matrix}x+6=2x+9\\x+6=-2x-9\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-2x=9-6\\x+2x=-9-6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}-3x=3\\3x=-15\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=-5\end{matrix}\right.\)
d) \(\left|2x-3\right|+x=21\)
\(\Leftrightarrow\left|2x-3\right|=21-x\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=21-x\\2x-3=-21+x\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x+x=21+3\\2x-x=-21+3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}3x=24\\x=-18\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-18\end{matrix}\right.\)
LL
27 tháng 6 2019
Noob ơi, bạn phải đưa vào máy tính ý solve cái là ra x luôn, chỉ tội là đợi hơi lâu
27 tháng 6 2019
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
CW
24 tháng 6 2016
a) \(\left(2y-1\right)^{1000}-\left(3+y\right)^{1000}=0\)
\(\Rightarrow\left(2y-1\right)^{1000}=\left(3+y\right)^{1000}\)
\(\Rightarrow2y-1=3+y\)
\(2y-y=3+1\)
\(y=4\)
b) \(\left(x-\frac{2}{9}\right)^3=\left(\frac{2}{3}\right)^6\)
\(\left(x-\frac{2}{9}\right)^3=\left(\left(\frac{2}{3}\right)^2\right)^3\)
\(\Rightarrow x-\frac{2}{9}=\left(\frac{2}{3}\right)^2\)
\(x-\frac{2}{9}=\frac{4}{9}\)
\(x=\frac{2}{3}\)
c) \(\left(2x-1\right)^6=\left(2x-1\right)^8\)
\(\left(\left(2x-1\right)^3\right)^2=\left(\left(2x-1\right)^4\right)^2\)
\(\Rightarrow\left(2x-1\right)^3=\left(2x-1\right)^4\)
\(8x^3-1=16x^4-1\)
\(16x^4-8x^3=0\)
\(8x^3\left(2x-1\right)=0\)
Nếu \(8x^3=0\) thì \(x^3=0\Rightarrow x=0\)
Nếu \(2x-1=0\)thì \(2x=1\Rightarrow x=\frac{1}{2}\)
Vậy x=0 và x=1/2
2x2 - 18x + 6x -6 = 16 + 25
2x2 - 12x -47 =0
\(x=\pm\dfrac{\sqrt{130}+6}{2}\)
Hằng đẳng thức: \(a^2+2ab+b^2=\left(a+b\right)^2\)
Cách chứng minh: \(VT=\left(a^2+ab\right)+\left(ab+b^2\right)=a\left(a+b\right)+b\left(a+b\right)\\ =\left(a+b\right)\left(a+b\right)=\left(a+b\right)^2=VP\)
Áp dụng:
Kiểu đề 1: \(2x\left(x-9\right)+3\left(2x\right)-6=4^2+5^2\\ \Rightarrow2x^2-18x+6x-6=16+25\\ \Rightarrow2x^2-12x-47=0\\ \Rightarrow x^2-6x-\dfrac{47}{2}=0\\ \Rightarrow\left(x^2-2.x.3+3^2\right)-9-\dfrac{47}{2}=0\\ \Rightarrow\left(x-3\right)^2=\dfrac{65}{2}=\left(\dfrac{\pm\sqrt{130}}{2}\right)^2\\\)
\(\Rightarrow\left[{}\begin{matrix}x-3=\dfrac{\sqrt{130}}{2}\\x-3=\dfrac{-\sqrt{130}}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{6+\sqrt{130}}{2}\\x=\dfrac{6-\sqrt{130}}{2}\end{matrix}\right.\)
Kiểu đề 2: \(2x\left(x-9\right)+3\left(2x-6\right)=4^2+5^2\\ \Rightarrow2x^2-18x+6x-18=16+25\\ \Rightarrow2x^2-12x-59=0\\ \Rightarrow x^2-6x-\dfrac{59}{2}=0\\ \Rightarrow\left(x^2-2.x.3+3^2\right)-9-\dfrac{59}{2}=0\\ \Rightarrow\left(x-3\right)^2=\dfrac{77}{2}=\left(\dfrac{\pm\sqrt{154}}{2}\right)^2\\ \)
\(\Rightarrow\left[{}\begin{matrix}x-3=\dfrac{\sqrt{154}}{2}\\x-3=\dfrac{-\sqrt{154}}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{6+\sqrt{154}}{2}\\x=\dfrac{6-\sqrt{154}}{2}\end{matrix}\right.\)