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\(x+15\%x=115\)
\(\Leftrightarrow x+\frac{15}{100}x=115\)
\(\Leftrightarrow x+\frac{3}{20}x=115\)
\(\Leftrightarrow x\left(1+\frac{3}{20}\right)=115\)
\(\Leftrightarrow x\left(\frac{20}{20}+\frac{3}{20}\right)=115\)
\(\Leftrightarrow\frac{23}{20}x=115\)
\(\Leftrightarrow x=115\div\frac{23}{20}\)
\(\Leftrightarrow x=115\times\frac{20}{23}\)
\(\Leftrightarrow x=\frac{2300}{23}=100\)

\(Tìmx:\)
\(a.x:3\frac{1}{15}=1\frac{1}{12}.\)
\(x:\frac{46}{15}=\frac{13}{12}.\)
\(x=\frac{13}{12}.\frac{46}{15}.\)
\(x=\frac{299}{90}.\)
\(Tính:\)
\(a.\frac{41}{42}+\frac{55}{56}+\frac{71}{72}+\frac{89}{90}BCNN=2520.\) \(=\frac{2460}{2520}+\frac{2475}{2520}+\frac{2485}{2520}+\frac{2492}{2520}.\)
\(=\frac{9912}{2520}.\)
\(=\frac{59}{15}.\)
CÂU B CỦA BÀI TÌM X MÌNH CHƯA HỌC NÊN KO BÍT LÀM!
a) x : \(3\frac{1}{5}=1\frac{1}{12}\)
x : \(\frac{16}{5}=\frac{13}{12}\)
x = \(\frac{13}{12}.\)\(\frac{16}{5}\)
x = \(\frac{208}{60}=\frac{52}{15}\)
b) x + 15% . x = 115
x + \(\frac{3}{20}\).x = 115
x.1 + \(\frac{3}{20}.x=115\)
x. (1+\(\frac{3}{20}\))=115
x. \(\frac{23}{20}=115\)
x = \(115:\frac{23}{20}\)
x=100

a) \(\left(x-47\right)-115=0\)\(\Leftrightarrow x-47=115\)\(\Leftrightarrow x=47+115=162\).
b) \(315+\left(146-x\right)=401\)\(\Leftrightarrow146-x=401-315\)\(\Leftrightarrow146-x=86\)\(\Leftrightarrow x=146-86\)\(\Leftrightarrow x=60\).

\(a,2016-\left|-16\right|=50-2x\)
\(2016-16=50-2x\)
\(2x=50-2016+16\)
\(2x=-1950\)
\(x=-975\)
Vậy \(x=-975\)
\(b,27-\left|32-x\right|=-\left(-5\right)\)
\(27-\left|32-x\right|=5\)
\(\left|32-x\right|=27-5\)
\(\left|32-x\right|=22\)
\(\Leftrightarrow\)\(\left[\begin{matrix}32-x=22\\32-x=-22\end{matrix}\right.\) \(\Leftrightarrow\) \(\left[\begin{matrix}x=10\\n=54\end{matrix}\right.\)
Vậy \(x\in\left\{10;54\right\}\)
\(c,-\left(83-115\right)-x=90\)
\(-\left(-32\right)-x=90\)
\(32-x=90\)
\(x=32-90\)
\(x=-58\)
Vậy \(x=-58\)

a, Câu hỏi của Nguyễn Ánh Ngân - Toán lớp 6 - Học toán với OnlineMath
b, Câu hỏi của Vũ Xuân Hiếu - Toán lớp 6 | Học trực tuyến
c)x−12−16−112−120−130−142−156=524
\(<=> x=\dfrac{5}{24}+\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\)
\(<=> x= \dfrac{13}{12}\)

a./ \(\Leftrightarrow x^{10}=1\Leftrightarrow x=\pm1\)
b./ \(\Leftrightarrow x^{10}-x=0\Leftrightarrow x\left(x^9-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x^9=1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)
c./ \(\Leftrightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\Leftrightarrow\left(2x-15\right)^3\left(\left(2x-15\right)^2-1\right)=0\Leftrightarrow\orbr{\begin{cases}2x-15=0\\\left(2x-15\right)^2=1\end{cases}}\)
- 2x - 15 = 0 \(\Leftrightarrow x=\frac{15}{2}\)
- 2x - 15 = 1 \(\Leftrightarrow x=\frac{16}{2}=8\)
- 2x - 15 = -1 \(\Leftrightarrow x=\frac{14}{2}=7\)

Ta có: x∈BC(9;8) và x nhỏ nhất
=> x∈BCNN(9;8)
9=32
8=23
BCNN(9;8)=23.32=72

\(B=\left(1+\dfrac{1}{8}\right)\left(1+\dfrac{1}{15}\right)\left(1+\dfrac{1}{24}\right).....\left(1+\dfrac{1}{440}\right)\left(1+\dfrac{1}{483}\right)\)
\(B=\dfrac{9}{8}.\dfrac{16}{15}.\dfrac{25}{24}.....\dfrac{441}{440}.\dfrac{484}{483}\)
\(B=\dfrac{9.16.25.....441.484}{8.15.24.....440.483}\)
\(B=\dfrac{3.3.4.4.5.5.....21.21.22.22}{2.4.3.5.4.6.....20.22.21.23}\)
\(B=\dfrac{3.4.5.....21.22}{2.3.4.....20.21}.\dfrac{3.4.5.....21.22}{4.5.6.....22.23}\)
\(B=11.\dfrac{3}{23}=\dfrac{33}{23}\)
B = \(\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}.\dfrac{25}{24}...\dfrac{121}{120}.\dfrac{144}{143}\)
B = \(\dfrac{4.9.16.25...121.144}{3.8.15.24....120.143}\)
B = \(\dfrac{2.2.3.3.4.4.5.5...11.11.12.12}{1.3.2.4.3.5.4.6...10.12.11.13}\)
B = \(\dfrac{2.3.4.5...11.12}{1.2.3.4.5...10.11}.\dfrac{2.3.4.5...11.12}{3.4.5.6.7...12.13}\)
B = 12 . \(\dfrac{2}{13}\)
B = \(\dfrac{24}{13}\)
\(\Leftrightarrow x+\frac{3}{20}x=115\)
\(\Rightarrow x\left(1+\frac{3}{20}\right)=115\)
\(\Rightarrow x\cdot\frac{23}{20}=115\)
\(\Rightarrow\frac{23x}{20}=115\Rightarrow115\cdot20=23x\)
=>2300=23x
=>x=2300:23
=>x=100
x(1+15%) = 115
x.23/20 = 115
x = 115 : 23/20 = 100