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Bài 2:
a: x+5/3=x-6/7
=>0x=-6/7-5/3(vô lý)
b: \(\Leftrightarrow x-10\left(\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}+...+\dfrac{2}{53\cdot55}\right)=\dfrac{3}{11}\)
\(\Leftrightarrow x-10\left(\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+...+\dfrac{1}{53}-\dfrac{1}{55}\right)=\dfrac{3}{11}\)
\(\Leftrightarrow x-10\cdot\dfrac{4}{55}=\dfrac{3}{11}\)
hay x=3/11+8/11=1
Bài 1:
a. $(-20)+x=-30$
$x-20=-30$
$x=-30+20=-(30-20)=-10$
b.
$(-10)-x=-20$
$x=(-10)-(-20)=-10+20=20-10=10$
c. Đề sai. Bạn xem lại.
d.
$x+(-3)=-7$
$x=-7-(-3)=-7+3=-(7-3)=-4$
e.
$x-(-5)=-9$
$x=(-9)+(-5)=-14$
f.
$x(-11)=12$
$x=\frac{12}{-11}=\frac{-12}{11}$
h.
$2x-10=20$
$2x=20+10=30$
$x=30:2=15$
l.
$4x-8=-8$
$4x=-8+8=0$
$x=0:4=0$
k.
$-12-(-2)x=-8$
$(-2)x=-12-(-8)=-12+8=-(12-8)=-4$
$x=(-4):(-2)=2$
Bài 2:
a. $-20-(10-x)=-3$
$10-x=-20-(-3)=-20+3=-(20-3)=-17$
$x=10-(-17)=10+17=27$
b.
$14+(14-x)=-2$
$14-x=-2-14=-16$
$x=14-(-16)=14+16=30$
c.
$-15-(x-3)=-7$
$x-3=-15-(-7)=-15+7=-8$
x=-8+3=-5$
d.
$(x+4)+(-20)=-8$
$x+4=-8-(-20)=-8+20=12$
$x=12-4=8$
e.
$-2x-2=-4$
$-2x=-4+2=-2$
$x=(-2):(-2)=1$
f.
$-2x+4=-4$
$-2x=-4-4=-8$
$x=(-8):(-2)=4$
l.
$-12-(-2)x=-2-4=-6$
$(-2)x=-12-(-6)=-12+6=-6$
$x=(-6):(-2)=3$
\(x-\dfrac{20}{11.13}-\dfrac{20}{23.15}-....-\dfrac{20}{53.55}=\dfrac{3}{11}\)
\(x-\left(\dfrac{20}{11.13}+\dfrac{20}{13.15}+....+\dfrac{20}{53.55}\right)=\dfrac{3}{11}\)
\(x-10\left(\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+...+\dfrac{1}{53}-\dfrac{1}{55}\right)=\dfrac{3}{11}\)
\(x-10\left(\dfrac{1}{11}-\dfrac{1}{55}\right)=\dfrac{3}{11}\)
\(x-\dfrac{8}{11}=\dfrac{3}{11}\)
=> \(x=\dfrac{3}{11}+\dfrac{8}{11}=1\)
Ta có:
x-\(\frac{20}{11.13}\)-\(\frac{20}{13.15}\)-\(\frac{20}{15.17}\)-...-\(\frac{20}{53.57}\)=\(\frac{3}{11}\)
\(\Rightarrow\)\(\frac{20}{11.13}\)-\(\frac{20}{13.15}\)-\(\frac{20}{15.17}\)-...-\(\frac{20}{53.57}\)= x-\(\frac{3}{11}\)
\(\frac{1}{10}\).\((\frac{20}{11.13}.\frac{20}{13.15}.\frac{20}{15.17}...\frac{20}{53.57})\)= x-\(\frac{3}{11}\)
\(\frac{2}{11.13}\).\(\frac{2}{13.15}\).\(\frac{2}{15.17}\)...\(\frac{2}{53.57}\)= x-\(\frac{3}{11}\)
\(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}\)-\(\frac{1}{17}\)+...+\(\frac{1}{53}\)-\(\frac{1}{57}\)=x-\(\frac{3}{11}\)
\(\frac{1}{11}-\frac{1}{57}\)=x-\(\frac{3}{11}\)
\(\frac{46}{627}\)=x-\(\frac{3}{11}\)
x=\(\frac{46}{627}\)-\(\frac{3}{11}\)
Vậy x=\(\frac{-125}{627}\)
\(x-\frac{20}{11.13}-\frac{20}{13.15}-\frac{20}{15.17}-...-\frac{20}{53.55}=\frac{3}{11}\)
\(x-\left(\frac{20}{11.13}+\frac{20}{13.15}+\frac{20}{15.17}+...+\frac{20}{53.55}\right)=\frac{3}{11}\)
\(x-\left[10.\left(\frac{2}{11.13}+\frac{2}{13.15}+\frac{2}{15.17}+...+\frac{2}{53.55}\right)\right]=\frac{3}{11}\)
\(x-\left[10.\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+...+\frac{1}{53}-\frac{1}{55}\right)\right]=\frac{3}{11}\)
\(x-\left[10.\left(\frac{1}{11}-\frac{1}{55}\right)\right]=\frac{3}{11}\)
\(x-\left[\frac{10}{11}-\frac{10}{55}\right]=\frac{3}{11}\)
\(x-\left[\frac{10}{11}-\frac{2}{11}\right]=\frac{3}{11}\)
\(x-\frac{8}{11}=\frac{3}{11}\)
\(x=\frac{3}{11}+\frac{8}{11}=1\)
a, \(15:\left(x+2\right)=3\Leftrightarrow x+2=3\Leftrightarrow x=1\)
b, \(20:\left(x+1\right)=2\Leftrightarrow x+1=10\Leftrightarrow x=9\)
c, \(240:\left(x-5\right)=2^2.5^2-20=80\Leftrightarrow x-5=3\Leftrightarrow x=8\)
d, \(96-3\left(x+1\right)=42\Leftrightarrow3\left(x+1\right)=54\Leftrightarrow x+1=18\Leftrightarrow x=17\)
a) 15 : (x + 2) = 3
x + 2 = 15 : 3
x + 2 = 5
x = 5 – 2 = 3
b) 20 : (1 + x) = 2
1 + x = 20 : 2
1 + x = 10
x = 10 – 1 = 9
c) 240 : (x – 5) = 22.52 – 20
240 : (x – 5) = 4.25 – 20
240 : (x - 5) = 100 – 20
240 : (x - 5) = 80
x – 5 = 240 : 80
x – 5 = 3
x = 3 + 5 = 8
d) 96 - 3(x + 1) = 42
3(x + 1) = 96 – 42
3(x + 1) = 54
x + 1 = 54 : 3
x + 1 = 18
x = 18 – 1
x = 17
a) x+15 = 20-4x
=> x=1
b) 17-x=7-6x
=> x=-2
c) -12+x=5x-20
=> x=2
d) 4x-5=15-x
=> x=4
e) 9x-7=20-6x
=>x= \(\frac{9}{5}\)
g)2.(x-10)=3.(x-20)=x-4
=> x thuộc ∅
h) (x^2+2).(x-3) <0
=> x=3,...
Ta có :
\(x-\frac{20}{11.13}-\frac{20}{13.15}-\frac{20}{15.17}-...-\frac{20}{53.55}=\frac{3}{1}\)
\(x-\left(\frac{20}{11.13}+\frac{20}{13.15}+\frac{20}{15.17}+...+\frac{20}{53.55}\right)=\frac{3}{1}\)
Đặt \(A=\frac{20}{11.13}+\frac{20}{13.15}+\frac{20}{15.17}+...+\frac{20}{53.55}\)
\(A=10\left(\frac{2}{11.13}+\frac{2}{13.15}+\frac{2}{15.17}+...+\frac{2}{53.55}\right)\)
\(A=10\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+...+\frac{1}{53}-\frac{1}{55}\right)\)
\(A=10\left(\frac{1}{11}-\frac{1}{55}\right)\)
\(A=10.\frac{4}{55}=\frac{40}{55}=\frac{8}{11}\)
\(\Rightarrow x-\frac{8}{11}=\frac{3}{1}=3\)
\(x=3+\frac{8}{11}=\frac{33}{11}+\frac{8}{11}=\frac{41}{11}\)
Ủng hộ mk nha !!! ^_^
\(x-3=20\\ x=20+3\\ x=23\)
Vậy \(x=23\)
|x-3|=20
=>\(\left[{}\begin{matrix}x-3=20\\x-3=-20\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=23\\x=-17\end{matrix}\right.\)