a,(2+4+6+...+2018)-2.x=2020
b,x.(x-5)=14
c,x.(2x+1)=21
(đăng cho các CTV thử việc ^-^)
Tìm x:
a, -5/6 - x = 7/12 + -1/3
b, (4,5 - 2x).(-1 4/7) = 11/14
c, (2/11 + 1/3).x = (1/7 - 1/8).56
`a)-5/6-x=7/12+[-1]/3`
`-5/6-x=1/4`
`x=-5/6-1/4`
`x=-13/12`
`b)(4,5-2x).(-1 4/7)=11/14`
`(4,5-2x).(-11/7)=11/14`
`4,5-2x=11/14:(-11/7)=-0,5`
`2x=4,5-(-0,5)=5`
`x=5:2=2,5`
`c)(2/11+1/3)x=(1/7-1/8).56`
`17/33 x=1/56 .56=1`
`x=33/17`
a.1/3 x 2019/2020
b. 2017 x 2018/2019 x 2019/2018
Có bao nhiêu giá trị của m trên [-2018; 2018] để phương trình
x2 + (2 - m)x + 4 = 4\(\sqrt{x^3+4x}\) có nghiệm ?
A. 2020
B. 2021
C. 2018
D. 2019
\(x=0\) không là nghiệm của phương trình
Chia hai vế phương trình cho x, phương trình trở thành:
\(\left(x+\dfrac{4}{x}\right)+2-m=4\sqrt{x+\dfrac{4}{x}}\left(1\right)\)
Đặt \(x+\dfrac{4}{x}=t\left(t\ge2\right)\)
\(\left(1\right)\Leftrightarrow m=f\left(t\right)=t^2-4t+2\left(2\right)\)
Phương trình đã cho có nghiệm khi phương trình \(\left(2\right)\) có nghiệm \(t\ge2\)
\(\Leftrightarrow m\ge f\left(2\right)=-2\)
\(\Rightarrow\) có 2021 giá trị thỏa mãn yêu cầu bài toán
a)|-x+2/5|+1/2=3,5 b)21/5+3:|x/4-2/3|=6
c)7,5-3|5-2x|=-4,5 d)1/3-|5/4-2x|=1/4
e)21/5+3:|x/4-2/3|=6
a)|-x+2/5|+1/2=3,5 b)21/5+3:|x/4-2/3|=6
c)7,5-3|5-2x|=-4,5 d)1/3-|5/4-2x|=1/4
e)21/5+3:|x/4-2/3|=6
a: Ta có: \(\left|\dfrac{2}{5}-x\right|+\dfrac{1}{2}=3.5\)
\(\Leftrightarrow\left|x-\dfrac{2}{5}\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{2}{5}=3\\x-\dfrac{2}{5}=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{17}{5}\\x=-\dfrac{13}{5}\end{matrix}\right.\)
b: Ta có: \(\dfrac{21}{5}+3:\left|\dfrac{x}{4}-\dfrac{2}{3}\right|=6\)
\(\Leftrightarrow3:\left|\dfrac{1}{4}x-\dfrac{2}{3}\right|=6-\dfrac{21}{5}=\dfrac{9}{5}\)
\(\Leftrightarrow\left|\dfrac{1}{4}x-\dfrac{2}{3}\right|=\dfrac{5}{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{4}x-\dfrac{2}{3}=\dfrac{5}{3}\\\dfrac{1}{4}x-\dfrac{2}{3}=-\dfrac{5}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{4}x=\dfrac{7}{3}\\\dfrac{1}{4}x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{28}{3}\\x=-4\end{matrix}\right.\)
Giải các phương trình sau
a) 22-x(1-4x)=(2x+3)^3
b) 2x/3 + 2x-1/6 = 4- x/3
c) x-1/2019 + x-2/2018 = x-3/2017 + x-4/2016
d) 2-x/2001 - 1 = 1-x/2002 - x/2003
e) 150-x/25 + 188-x/21 + 201-x/19 +171-x/23 =0
a) \(22-x\left(1-4x\right)=\left(2x+3\right)^3\)
\(\Leftrightarrow22-x+4x^2=8x^3+36x^2+54x+27\)
\(\Leftrightarrow-x-54x+4x^2-36x^2-8x^3=-22+27\)
\(\Leftrightarrow-8x^3-32x^2-55x=5\Leftrightarrow-8x^3-32x^2-55x-5=0\)
Bn tự làm tiếp nhé
b) \(\frac{2x}{3}+\frac{2x-1}{6}=\frac{4-x}{3}\Leftrightarrow\frac{2.2x}{6}+\frac{2x-1}{6}=\frac{2\left(4-x\right)}{6}\)
\(\Leftrightarrow2.2x+2x-1=2\left(4-x\right)\Leftrightarrow4x+2x-1=8-2x\)
\(\Leftrightarrow6x-1=8-2x\Leftrightarrow8x=9\Leftrightarrow x=\frac{9}{8}\)
Vậy phương trình có tập nghiệm S ={9/8}
c) \(\frac{x-1}{2019}+\frac{x-2}{2018}=\frac{x-3}{2017}+\frac{x-4}{2016}\)
\(\Leftrightarrow\left(\frac{x-1}{2019}-1\right)+\left(\frac{x-2}{2018}-1\right)=\left(\frac{x-3}{2017}-1\right)+\left(\frac{x-4}{2016}-1\right)\)
\(\Leftrightarrow\frac{x-2020}{2019}+\frac{x-2020}{2018}-\frac{x-2020}{2017}-\frac{x-2020}{2016}=0\)
\(\Leftrightarrow\left(x-2020\right)\left(\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}\right)=0\)
Do \(\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}>0\)
Nên \(x-2020=0\Leftrightarrow x=2020\)
a, (x+2018)(1/2+2/7)=(x+2018).(1/5+1/6)
b, 7(x-1)+2x(x-1)=0
a) \(\left(x+2018\right)\left(\frac{1}{2}+\frac{2}{7}\right)=\left(x+2018\right)\left(\frac{1}{5}+\frac{1}{6}\right)\)
\(\Leftrightarrow\) \(\left(x+2018\right)\left(\frac{1}{2}+\frac{2}{7}\right)-\left(x+2018\right)\left(\frac{1}{5}+\frac{1}{6}\right)\) = 0
\(\Leftrightarrow\left(x+2018\right)\left(\frac{1}{2}+\frac{2}{7}-\frac{1}{5}-\frac{1}{6}\right)=0\)
\(\Leftrightarrow x+2018=0\)
\(\Leftrightarrow x=-2018\)
b) \(7\left(x-1\right)+2x\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(7+2x\right)=0\)
\(\Leftrightarrow\) x - 1 = 0 hoặc 7 + 2x = 0
1) x - 1 = 0 \(\Leftrightarrow\) x = 1
2) 7 + 2x = 0 \(\Leftrightarrow\) -3,5
Vậy: x = 1; -3,5
b) \(7\left(x-1\right)+2x\left(x-1\right)=0\)
=> \(\left(x-1\right).\left(7+2x\right)=0\)
=> \(\left\{{}\begin{matrix}x-1=0\\7+2x=0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x=0+1\\2x=0-7=-7\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x=1\\x=\left(-7\right):2\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=1\\x=-\frac{7}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{1;-\frac{7}{2}\right\}.\)
Chúc bạn học tôt!
tìm x
a, | 5/4x - 7/2 | - | 5/8x + 3/5 | = 0
b, 21/5 + 3 : | x/4 - 2/3 | = 6
c,| 9 + x | = 2x
d, | 2x - 3 | + x = 21
e, | 7 - 2x | + 7 = 2x
f, | -x + 2/5 | + 1/2 = 3,5
G, | 3x - 4 | + 4 = 3x
a) | 5/4x -7/2| - | 5/8x + 3/5| = 0
|5/4x - 7/2| = | 5/8x + 3/5|
TH1: 5/4x - 7/2 = 5/8x + 3/5
=> 5/4x - 5/8x = 3/5 +7/2
5/8x = 41/10
x = 41/10:5/8
x = 164/25
TH2: 5/4x - 7/2 = -5/8x - 3/5
=> 5/4x + 5/8x = -3/5 +7/2
15/8x = 29/10
x = 29/10 : 15/8
x = 116/75
KL: x = 164/25 hoặc x = 116/75
các bài cn lại b lm tương tự nha! h lm dài lắm!
tìm x
a, | 5/4x - 7/2 | - | 5/8x + 3/5 | = 0
b, 21/5 + 3 : | x/4 - 2/3 | = 6
c,| 9 + x | = 2x
d, | 2x - 3 | + x = 21
e, | 7 - 2x | + 7 = 2x
f, | -x + 2/5 | + 1/2 = 3,5
G, | 3x - 4 | + 4 = 3x
a: =>|5/4x-7/2|=|5/8x+3/5|
=>5/4x-7/2=5/8x+3/5 hoặc 5/4x-7/2=-5/8x-3/5
=>5/8x=41/10 hoặc 15/8x=29/10
=>x=164/25 hoặc x=116/75
b: =>3:|x/4-2/3|=6-21/5=9/5
=>|1/4x-2/3|=5/3
=>1/4x-2/3=5/3 hoặc 1/4x-2/3=-5/3
=>1/4x=7/3 hoặc 1/4x=-1
=>x=28/3 hoặc x=-4
c: \(\Leftrightarrow\left\{{}\begin{matrix}x>=0\\\left(2x-x-9\right)\left(2x+x+9\right)=0\end{matrix}\right.\Leftrightarrow x=9\)
e: =>|2x-7|=2x-7
=>2x-7>=0
=>x>=7/2