A)X-2/2x-2<1/2
B)3x^2-x-10=0
C) x+1/2x-3>1
Biểu thức: 21-(2x-4)(x+1) bằng: A. 21-(4-2x).(x+1) B. 21+(4-2x).(x+1) C .21+(4-2x).(x-1)
Câu 29: Biểu thức : 21 - (2x - 4)(x + 1) bằng: A. 21 -(4- 2x)(x + 1) B. 21 + (4 -2x)(x + 1) C. 21 +(4- 2x)(x – 1)
\(21-\left(2x-4\right)\left(x+1\right)\)
\(=21-2x^2-2x+4x+4\)
\(=-2x^2+2x+25\)
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: \(\dfrac{21}{5}+3:\left|\dfrac{x}{4}-\dfrac{2}{3}\right|=6\)
\(\Leftrightarrow3:\left|\dfrac{1}{4}x-\dfrac{2}{3}\right|=6-\dfrac{21}{5}=\dfrac{9}{5}\)
\(\Leftrightarrow\left|\dfrac{1}{4}x-\dfrac{2}{3}\right|=\dfrac{5}{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{4}x-\dfrac{2}{3}=\dfrac{5}{3}\\\dfrac{1}{4}x-\dfrac{2}{3}=-\dfrac{5}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{4}x=\dfrac{7}{3}\\\dfrac{1}{4}x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{28}{3}\\x=-4\end{matrix}\right.\)
a) |9+x|=2x b) |5x|-3x=2 c) |x+6|-9=2x d)|2x-3|+x=21
a, \(\left|9+x\right|=2x\)
\(\Leftrightarrow\orbr{\begin{cases}9+x=2x\\9+x=-2x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}9=2x-x\\9=-2x-x\end{cases}\Leftrightarrow\orbr{\begin{cases}9=x\\9=-3x\end{cases}}}\)
\(\Rightarrow\orbr{\begin{cases}x=9\\x=-3\end{cases}}\)
b, \(\left|5x\right|-3x=2\)
\(\Leftrightarrow\orbr{\begin{cases}5x-3x=2\\5x-3x=-2\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=2\\2x=-2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)
c, \(\left|x+6\right|-9=2x\)
\(\Leftrightarrow\orbr{\begin{cases}x+6-9=2x\\x+6-9=-2x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x+6=2x+9\\x+6=-2x+9\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x+6=2x+9\\x+6=7x\end{cases}\Leftrightarrow}\orbr{\begin{cases}x+6=3\\x=7x-6\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-3\\x=1\end{cases}}\)
d. Tương tự
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!
a.( x - 2) chia hết cho 32 , ( x - 2 ) chia hết cho 48 và 0 < x < 100
b. ( x - 62 ) = 90 . 21 chia hết cho( 2x + 5 )
c) 21 chia hết ( 2x +5 )
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
a) + + b) + - c) - + d) - -
a \(\dfrac{1}{x-y}+\dfrac{2}{x+y}+\dfrac{3x}{y^2-x^2}\)
\(=\dfrac{x+y+2x-2y-3x}{\left(x-y\right)\left(x+y\right)}=\dfrac{-y}{\left(x-y\right)\left(x+y\right)}\)
b: \(\dfrac{1}{x-2}+\dfrac{1}{x+2}-\dfrac{4x-4}{x^2-4}\)
\(=\dfrac{x+2+x-2-4x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{-2x+4}{\left(x-2\right)\left(x+2\right)}\)
=-2/x+2
c: \(\dfrac{x+1}{x+3}-\dfrac{x-1}{3-x}+\dfrac{2x-2x^2}{x^2-9}\)
\(=\dfrac{\left(x+1\right)\left(x-3\right)+\left(x-1\right)\left(x+3\right)+2x-2x^2}{\left(x+3\right)\left(x-3\right)}\)
\(=\dfrac{x^2-2x-3+x^2+2x-3+2x-2x^2}{\left(x+3\right)\left(x-3\right)}\)
\(=\dfrac{2x-6}{\left(x+3\right)\left(x-3\right)}=\dfrac{2}{x+3}\)