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a) x−(−15)=−13−(−85−13)
=> x+15=85
=>x=70
b) (−9−x)+(x−14)=17−(−8+x)
=>-9-x+x-14=17+8-x
=>-23=25-x
=>x=48
Bài 7:
a, \(x\) = \(\dfrac{1}{5}\) + \(\dfrac{2}{11}\)
\(x\) = \(\dfrac{11}{55}\) + \(\dfrac{10}{55}\)
\(x=\dfrac{21}{55}\)
b, \(\dfrac{x}{15}\) = \(\dfrac{3}{5}\) - \(\dfrac{2}{3}\)
\(\dfrac{x}{15}\) = \(\dfrac{9}{15}\) - \(\dfrac{10}{15}\)
\(\dfrac{x}{15}\) = \(\dfrac{1}{15}\)
\(x\) = 1
c, \(\dfrac{11}{8}\) + \(\dfrac{13}{6}\)= \(\dfrac{85}{x}\)
\(\dfrac{33}{24}\) + \(\dfrac{52}{24}\) = \(\dfrac{85}{x}\)
\(\dfrac{85}{24}\) = \(\dfrac{85}{x}\)
24 = \(x\)
Giải:
Đặt \(A=\dfrac{6}{4.7}+\dfrac{6}{7.10}+\dfrac{6}{10.13}+...+\dfrac{6}{82.85}\)
\(\dfrac{1}{2}A=\dfrac{3}{4.7}+\dfrac{3}{7.10}+\dfrac{3}{10.13}+...+\dfrac{3}{82.85}\)
\(\dfrac{1}{2}A=\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+...+\dfrac{1}{82}-\dfrac{1}{85}\)
\(\dfrac{1}{2}A=\dfrac{1}{4}-\dfrac{1}{85}\)
\(\dfrac{1}{2}A=\dfrac{81}{340}\)
\(\Rightarrow A=\dfrac{81}{340}:\dfrac{1}{2}\)
\(A=\dfrac{81}{170}\)
mình viết đáp án luôn nhé
a,x=845/528
b,-7/5
c,-47/140
d,43/1710
chúc bạn làm bài thành công nhe!!@
các bạn nhớ chọn k đúng cho mình nhé
b: \(\Leftrightarrow x-15-27-x+x-13=-1\)
\(\Leftrightarrow x-55=-1\)
hay x=54
a) Ta có: x-43=(35-x)-48
⇔x-43=35-x-48
⇔x-43=-x-13
⇔x-43+x+13=0
⇔2x-30=0
⇔2(x-15)=0
mà 2≠0
nên x-15=0
hay x=15
b) Ta có: 305-x+14=48+(x-23)
⇔319-x=48+x-23
⇔319-x=x+25
⇔319-x-x-25=0
⇔-2x+294=0
⇔-2x=-294
hay x=147
Vậy: x=147
c) Ta có: -(x-6+85)=(x+51)-54
⇔-x+6-85=x+51-54
⇔-x-79=x-3
⇔-x-79-x+3=0
⇔-2x-76=0
⇔-2x=76
hay x=-38
Vậy: x=-38
d) Ta có: -(35-x)-(37-x)=33-x
⇔-35+x-37+x-33+x=0
⇔3x-105=0
⇔3(x-35)=0
mà 3≠0
nên x-35=0
hay x=35
Vậy: x=35
e) Ta có: 13-|x|=|-4|
⇔13-|x|=4
⇔|x|=9
⇔x∈{9;-9}
Vậy: x∈{9;-9}
f) Ta có: |x|-3+6=16
⇔|x|+3=16
⇔|x|=13
hay x∈{-13;13}
Vậy: x∈{-13;13}
g) Ta có: 35-|2x-1|=14
⇔|2x-1|=21
⇔\(\left[{}\begin{matrix}2x-1=21\\2x-1=-21\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=22\\2x=-20\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-10\end{matrix}\right.\)
Vậy: x∈{11;-10}
h) Ta có: |3x-2|+5=9-x
⇔|3x-2|+5-9+x=0
⇔|3x-2|-4+x=0
⇔|3x-2|=0-(-4+x)
⇔|3x-2|=4-x
⇔\(\left[{}\begin{matrix}3x-2=4-x\\3x-2=x-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x-2-4+x=0\\3x-2-x+4=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}4x-6=0\\2x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=6\\2x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{3}{2}\left(ktm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)
Vậy: x=-1
\(\frac{11}{8}+\frac{13}{6}=\frac{85}{x}\)
\(< =>\frac{66}{48}+\frac{104}{48}=\frac{85}{x}\)
\(< =>\frac{170}{48}=\frac{85}{x}\)
\(< =>170x=85.48=4080\)
\(< =>x=\frac{4080}{170}=24\)
11/8+13/6=85/x
85/x=66/48+104/48
85/x=170/48=85/20
suy ra:x=20
vậy x=20