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a)15/8+3/4-5/12
=45+18-10/24
=53/24
b)11/24.12/33+5/6
=11.12/12.2.11.3+5/6
=1/6+5/6
=6/6=1
c)15/8+7/24:5/8
=15/8+7/24.8/5
=15/8+7.8/3.8.5
=15/8+7/15
=đề sai, nếu đúng thì như này
=8/15+7/15
=15/15=1
\(\frac{8}{12}\cdot x\cdot4-\frac{18}{12}\cdot x\cdot6=\frac{5}{12}\)
\(x\left(\frac{8}{12}\cdot4-\frac{18}{12}\cdot6\right)=\frac{5}{12}\)
\(x\left(\frac{32}{12}-\frac{108}{12}\right)=\frac{5}{12}\)
\(x\left(-\frac{19}{3}\right)=\frac{5}{12}\)
\(x=\frac{5}{12}\div\left(-\frac{19}{3}\right)=-\frac{5}{76}\)
Vậy \(x=-\frac{5}{76}\)
Chúc bạn học tốt ^^!!!
Bài 46:
11: Ta có: \(-4\left|x-2\right|=-8\)
\(\Leftrightarrow\left|x-2\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=2\\x-2=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=0\end{matrix}\right.\)
Vậy: x∈{0;4}
12: Ta có: \(5\left|x+2\right|=-10\cdot\left(-2\right)\)
\(\Leftrightarrow5\left|x+2\right|=20\)
\(\Leftrightarrow\left|x+2\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-6\end{matrix}\right.\)
Vậy: x∈{-6;2}
13: Ta có: \(6\left|x-2\right|=18:\left(-3\right)\)
\(\Leftrightarrow6\left|x-2\right|=-6\)(1)
Ta có: \(\left|x-2\right|\ge0\forall x\)
\(\Rightarrow6\left|x-2\right|\ge0\forall x\)(2)
Ta có: -6<0(3)
Từ (1), (2) và (3) suy ra x∈∅
Vậy: x∈∅
14: Ta có:\(-7\left|x+4\right|=21:\left(-3\right)\)
\(\Leftrightarrow-7\left|x+4\right|=-7\)
\(\Leftrightarrow\left|x+4\right|=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=1\\x+4=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-5\end{matrix}\right.\)
Vậy: x∈{-5;-3}
15: Ta có: \(4\left|x+1\right|=8\left(-2\right)-8\left(-5\right)\)
\(\Leftrightarrow4\left|x+1\right|=-16-\left(-40\right)\)
\(\Leftrightarrow4\left|x+1\right|=24\)
\(\Leftrightarrow\left|x+1\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=6\\x+1=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-7\end{matrix}\right.\)
Vậy: x∈{-7;5}
16: Ta có: \(3\left|x+5\right|=-9\)(4)
Ta có: |x+5|≥0∀x
⇒3|x+5|≥0∀x(5)
Ta có: -9<0(6)
Từ (4), (5) và (6) suy ra x∈∅
Vậy: x∈∅
17: Ta có: \(-8\left|x-3\right|=24-16:2\)
\(\Leftrightarrow-8\left|x-3\right|=16\)
\(\Leftrightarrow\left|x-3\right|=-2\)
mà |x-3|≥0>-2∀x
nên x∈∅
Vậy: x∈∅
18: Ta có: \(-3\left|x+6\right|=6\cdot2-9\)
\(\Leftrightarrow-3\left|x+6\right|=3\)
\(\Leftrightarrow\left|x+6\right|=-1\)
mà |x+6|≥0>-1∀x
nên x∈∅
Vậy: x∈∅
19: Ta có: \(5-\left|x+7\right|=4\)
\(\Leftrightarrow\left|x+7\right|=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x+7=-1\\x+7=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=-6\end{matrix}\right.\)
Vậy: x∈{-8;-6}
20: Ta có: \(12-\left|x+8\right|=10\)
\(\Leftrightarrow\left|x+8\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x+8=2\\x+8=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=-10\end{matrix}\right.\)
Vậy: x∈{-10;-6}
\(\frac{1}{6}=\frac{x}{18}\Rightarrow x=3\)
\(\frac{x}{8}=-\frac{1}{4}\Rightarrow x=-2\)
\(\frac{4}{-5}=\frac{x}{10}\Rightarrow x=-8\)
\(\frac{11}{5}=-\frac{22}{x}\Rightarrow x=-10\)
\(\frac{x}{8}=\frac{8}{x}\Rightarrow x^2=64\Rightarrow x=8\)
\(\frac{x}{-11}=-\frac{11}{x}\Rightarrow x^2=121\Rightarrow x=11\)
#H
a: =7/8:(2/9-18+1/36)-5/12
=-7/142-5/12=-397/852
b: =3/7(4/9+5/9:6/12)=2/3
c: =5^8(16/31-47/31)+1/3=-5^8+1/3
d: =7/2(3/8+5/8:4/15)=609/64
a: \(\left(-\dfrac{5}{6}+\dfrac{2}{5}\right):\dfrac{3}{8}+\left(\dfrac{4}{5}-\dfrac{11}{30}\right):\dfrac{3}{8}\)
\(=\left(-\dfrac{5}{6}+\dfrac{2}{5}\right)\cdot\dfrac{8}{3}+\left(\dfrac{4}{5}-\dfrac{11}{30}\right)\cdot\dfrac{8}{3}\)
\(=\dfrac{8}{3}\left(-\dfrac{5}{6}+\dfrac{2}{5}+\dfrac{4}{5}-\dfrac{11}{30}\right)\)
\(=\dfrac{8}{3}\cdot\dfrac{-25+36-11}{30}\)
=0
b: \(\left(-\dfrac{3}{4}+\dfrac{2}{5}\right):\dfrac{3}{7}+\left(\dfrac{3}{5}+\dfrac{-1}{4}\right):\dfrac{3}{7}\)
\(=\left(-\dfrac{3}{4}+\dfrac{2}{5}\right)\cdot\dfrac{7}{3}+\left(\dfrac{3}{5}-\dfrac{1}{4}\right)\cdot\dfrac{7}{3}\)
\(=\dfrac{7}{3}\left(-\dfrac{3}{4}+\dfrac{2}{5}+\dfrac{3}{5}-\dfrac{1}{4}\right)\)
\(=\dfrac{7}{3}\cdot0=0\)
c: \(\dfrac{-13}{18}\cdot\dfrac{5}{8}+\dfrac{-5}{18}\cdot\dfrac{2}{9}+\dfrac{-13}{18}\cdot\dfrac{3}{8}+\dfrac{-5}{18}\cdot\dfrac{7}{9}\)
\(=\left(-\dfrac{13}{18}\cdot\dfrac{5}{8}+\dfrac{-13}{18}\cdot\dfrac{3}{8}\right)+\left(-\dfrac{5}{18}\cdot\dfrac{2}{9}+\dfrac{-5}{18}\cdot\dfrac{7}{9}\right)\)
\(=-\dfrac{13}{18}\left(\dfrac{5}{8}+\dfrac{3}{8}\right)+\dfrac{-5}{18}\left(\dfrac{2}{9}+\dfrac{7}{9}\right)\)
\(=-\dfrac{13}{18}-\dfrac{5}{18}=-\dfrac{18}{18}=-1\)
d: Sửa đề: \(\dfrac{-11}{19}\cdot\dfrac{4}{9}+\dfrac{-8}{19}\cdot\dfrac{3}{7}+\dfrac{-11}{19}\cdot\dfrac{5}{9}+\dfrac{-8}{19}\cdot\dfrac{4}{7}\)
\(=\left(-\dfrac{11}{19}\cdot\dfrac{4}{9}+\dfrac{-11}{19}\cdot\dfrac{5}{9}\right)+\left(\dfrac{-8}{19}\cdot\dfrac{3}{7}+\dfrac{-8}{19}\cdot\dfrac{4}{7}\right)\)
\(=-\dfrac{11}{19}\left(\dfrac{4}{9}+\dfrac{5}{9}\right)+\dfrac{-8}{19}\left(\dfrac{3}{7}+\dfrac{4}{7}\right)\)
\(=-\dfrac{11}{19}-\dfrac{8}{19}=-\dfrac{19}{19}=-1\)
\(a.\left(-\dfrac{5}{6}+\dfrac{2}{5}\right):\dfrac{3}{8}+\left(\dfrac{4}{5}-\dfrac{11}{30}\right):\dfrac{3}{8}\)
\(=\left(-\dfrac{13}{30}\right):\dfrac{3}{8}+\dfrac{13}{30}:\dfrac{3}{8}\)
\(=\left[\left(-\dfrac{13}{30}+\dfrac{13}{30}\right)\right]:\dfrac{3}{8}\)
\(=0:\dfrac{3}{8}=0\)
\(b.\left(-\dfrac{3}{4}+\dfrac{2}{5}\right):\dfrac{3}{7}+\left(\dfrac{3}{5}+-\dfrac{1}{4}\right):\dfrac{3}{7}\)
\(=\left(-\dfrac{7}{20}\right):\dfrac{3}{7}+\dfrac{7}{20}:\dfrac{3}{7}\)
\(=\left[\left(-\dfrac{7}{20}+\dfrac{7}{20}\right)\right]:\dfrac{3}{7}=0:\dfrac{3}{7}=0\)
\(c.-\dfrac{13}{18}.\dfrac{5}{8}+-\dfrac{5}{18}.\dfrac{2}{9}+-\dfrac{13}{18}.\dfrac{3}{8}+-\dfrac{5}{18}.\dfrac{7}{9}\)
\(=\left(\dfrac{5}{8}+\dfrac{3}{8}\right).-\dfrac{13}{18}+\left(\dfrac{2}{9}+\dfrac{7}{9}\right).-\dfrac{5}{18}\)
\(=1.-\dfrac{13}{18}+1.-\dfrac{5}{18}=-\dfrac{13}{18}+-\dfrac{5}{18}=-1\)
\(d.-\dfrac{11}{19}.\dfrac{4}{9}+\dfrac{-8}{19}.\dfrac{3}{7}+-\dfrac{11}{19}.\dfrac{5}{9}+-\dfrac{9}{19}.\dfrac{4}{7}\)
\(=\left(\dfrac{4}{9}+\dfrac{5}{9}\right).-\dfrac{11}{19}+-\dfrac{24}{133}+-\dfrac{36}{133}\)
\(=-\dfrac{11}{19}+-\dfrac{60}{133}=-\dfrac{137}{133}\)
\(\frac{x}{-4}=\frac{-16}{x}\)
\(\Rightarrow x\cdot x=-16\cdot\left(-4\right)\)
\(\Rightarrow x^2=64\)
\(\Rightarrow x^2=\left(\pm8\right)^2\)
\(\Rightarrow x=\pm8\)
a)\(\left(x-\frac{5}{8}\right).\frac{5}{18}=-\frac{15}{36}\)
\(\Rightarrow x-\frac{5}{8}=\frac{-15}{36}.\frac{18}{5}\)
\(\Rightarrow x-\frac{5}{8}=-\frac{3}{2}\)
\(\Rightarrow x=-\frac{12}{8}+\frac{5}{8}=-\frac{7}{8}\)
b)\(\frac{x}{-4}=\frac{-16}{x}\)
\(\Rightarrow x^2=64\)
\(\Rightarrow x=\orbr{\begin{cases}8\\-8\end{cases}}\)
\(x-\dfrac{4}{8}=\dfrac{5}{18}\\ x=\dfrac{5}{18}+\dfrac{4}{8}\\ x=\dfrac{5}{18}+\dfrac{1}{2}\\ x=\dfrac{5}{18}+\dfrac{9}{18}\\ x=\dfrac{5+9}{18}\\ x=\dfrac{14}{18}\\ x=\dfrac{7}{9}\)