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Câu 13 :
\(\left(-\frac{1}{4}+\frac{5}{8}\right)+-\frac{3}{5}\)
\(=\frac{3}{8}-\frac{-3}{5}\)
\(=\frac{39}{40}\)
Câu 14 :
\(M=\frac{5}{9}.\frac{7}{13}+\frac{5}{9}.\frac{9}{13}-\frac{5}{9}.\frac{3}{13}\)
\(=\frac{5}{9}.\left(\frac{7}{13}+\frac{9}{13}-\frac{3}{13}\right)\)
\(=\frac{5}{9}.1=\frac{5}{9}\)
Câu 15 :
\(E=\left(-\frac{3}{4}+\frac{2}{5}\right):\frac{3}{7}+\left(\frac{3}{5}+\frac{-1}{4}\right):\frac{3}{7}\)
\(E=\left(-\frac{3}{4}+\frac{2}{5}+\frac{3}{5}+\frac{-1}{4}\right):\frac{3}{7}\)
\(E=0\)
Câu 16 :
\(H=\frac{7}{8}:\left(\frac{2}{9}-\frac{1}{8}\right)+\frac{7}{8}:\left(\frac{1}{36}-\frac{5}{12}\right)\)
\(=\frac{7}{8}:\left(\frac{2}{9}-\frac{1}{8}+\frac{1}{36}-\frac{5}{12}\right)\)
\(=\frac{7}{8}:\frac{-7}{24}=-3\)
huhuhuhuhuhuhuhuhuhuhuhuhu giúp mk đi
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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
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}
1) |x + 2| = 4
\(\Leftrightarrow\orbr{\begin{cases}x+2=4\\x+2=-4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=-6\end{cases}}\)
2) 3 – |2x + 1| = (-5)
\(\Leftrightarrow\left|2x+1\right|=8\Leftrightarrow\orbr{\begin{cases}2x+1=8\\2x+1=-8\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{7}{2}\\x=\frac{-9}{2}\end{cases}}\)
3) 12 + |3 – x| = 9
\(\Leftrightarrow\left|3-x\right|=-3\)(vô lí)
=>\(x=\varnothing\)
1) I x+2 I=4
\(\Rightarrow\orbr{\begin{cases}x+2=4\\x+2=-4\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-6\end{cases}}}\)
2) \(3-|2x+1|=-5\)
\(\Leftrightarrow|2x+1|=8\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=8\\2x+1=-8\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=7\\2x=-9\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{7}{2}\\x=\frac{-9}{2}\end{cases}}}\)
3) \(12+|3-x|=9\)
\(\Leftrightarrow|3-x|=-3\)(vô lí vì I 3-x I \(\ge\)0)
dễ mk bn cho mình hỏi nhé câu 4 là \(\frac{1}{2\cdot3}\)hay là\(\frac{1}{2}\cdot3\)
mình đang cần gấp câu trả lời ,bạn nào giải được nhanh k luôn ,hứa đấy
a) \(\frac{2}{5}+x=\frac{3}{4}\)
\(x=\frac{3}{4}-\frac{2}{5}\)
\(x=\frac{15}{20}-\frac{8}{20}\)
\(x=\frac{7}{20}\)
\(\)b)
\(x-\frac{1}{15}=\frac{3}{10}\\ x=\frac{3}{10}+\frac{1}{15}\\ x=\frac{9}{30}+\frac{2}{30}\\ x=\frac{11}{30}\)
c)
\(\frac{9}{8}-x=\frac{5}{12}\\ x=\frac{9}{8}-\frac{5}{12}\\ x=\frac{27}{24}-\frac{10}{24}\\ x=\frac{17}{24}\)
d)
\(\frac{3}{5}+x=\frac{5}{4}+\frac{7}{10}\\ \frac{3}{5}+x=\frac{25}{20}+\frac{14}{20}\\\frac{3}{5}+x=\frac{39}{20}\\ x=\frac{39}{20}-\frac{3}{5}\\ x=\frac{39}{20}-\frac{12}{20}\\ x=\frac{27}{20} \)
e)
\(\frac{9}{8}-x=\frac{3}{20}+\frac{2}{5}\\ \frac{9}{8}-x=\frac{3}{20}+\frac{8}{20}\\ \frac{9}{8}-x=\frac{11}{20}\\ x=\frac{9}{8}-\frac{11}{20}\\ x=\frac{45}{40}-\frac{22}{40}\\ x=\frac{23}{40}\)
g)
\(x+\frac{1}{3}=\frac{5}{6}+1\frac{7}{10}\\ x+\frac{1}{3}=\frac{5}{6}+\frac{17}{10}\\ x+\frac{1}{3}=\frac{25}{30}+\frac{51}{30}\\ x+\frac{1}{3}=\frac{76}{30}=\frac{38}{15}\\ x=\frac{38}{15}-\frac{1}{3}\\ x=\frac{38}{15}-\frac{5}{15}\\ x=\frac{33}{15}=\frac{11}{5}\)
h)
\(3x-\frac{3}{5}=\frac{1}{2}\\ 3x=\frac{1}{2}+\frac{3}{5}\\ 3x=\frac{5}{10}+\frac{6}{10}\\ 3x=\frac{11}{10}\\ x=\frac{11}{10}:3\\ x=\frac{11}{10}\cdot\frac{1}{3}\\ x=\frac{11}{30}\)
i)
\(4x+\frac{5}{12}+\frac{4}{9}=1\frac{13}{18}\\ 4x+\frac{5}{12}+\frac{4}{9}=\frac{31}{18}\\ 4x+\frac{5}{12}=\frac{31}{18}-\frac{4}{9}\\ 4x+\frac{5}{12}=\frac{31}{18}-\frac{8}{18}\\ 4x+\frac{5}{12}=\frac{23}{18}\\ 4x=\frac{23}{18}-\frac{5}{12}\\ 4x=\frac{46}{36}-\frac{15}{36}\\ 4x=\frac{31}{36}\\ x=\frac{31}{36}:4\\ x=\frac{31}{36}\cdot\frac{1}{4}\\ x=\frac{31}{144}\)
k)
\(2-\left(3x+\frac{3}{7}\right)=\frac{9}{21}\\ 2-\left(3x+\frac{3}{7}\right)=\frac{3}{7}\\3x+\frac{3}{7}=2-\frac{3}{7}\\ 3x+\frac{3}{7}=\frac{14}{7}-\frac{3}{7}\\ 3x+\frac{3}{7}=\frac{11}{7}\\ 3x=\frac{11}{7}-\frac{3}{7}\\ 3x=\frac{8}{7}\\ x=\frac{8}{7}:3\\ x=\frac{8}{7}\cdot\frac{1}{3}\\ x=\frac{8}{21} \)
Câu 5 nữa bạn cố gắng làm giúp mik nha ko thì mình 0 điểm á, nha iu nhìu lém