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a) x= 1/4
b) x= -7/144
c) x= 25/6
d) x=17/7
e) x= 80/63
f) x= -45/14
\(a.\)
\(\frac{4}{9}.x=\frac{1}{9}\)
\(\Rightarrow x=\frac{1}{9}:\frac{4}{9}\)
\(\Rightarrow x=\frac{1}{9}.\frac{9}{4}\)
\(\Rightarrow x=\frac{1}{4}\)
Vậy \(x=\frac{1}{4}\)
\(b.\)
\(\frac{7}{18}.x-\frac{2}{3}=\frac{5}{18}\)
\(\Rightarrow\frac{7}{18}.x=\frac{5}{18}+\frac{2}{3}\)
\(\Rightarrow\frac{7}{18}.x=\frac{5}{18}+\frac{12}{18}\)
\(\Rightarrow\frac{7}{18}.x=\frac{17}{18}\)
\(\Rightarrow x=\frac{17}{18}:\frac{7}{18}\)
\(\Rightarrow x=\frac{17}{18}.\frac{18}{7}\)
\(\Rightarrow x=\frac{17}{7}\)
Vậy \(x=\frac{17}{7}\)
\(c.\)
\(x:-\frac{1}{12}=\frac{7}{12}\)
\(\Rightarrow x=\frac{7}{12}.-\frac{1}{12}\)
\(\Rightarrow x=-\frac{7}{144}\)
Vậy \(x=-\frac{7}{144}\)
\(d.\)
\(\frac{4}{9}-\frac{7}{8}.x=-\frac{2}{3}\)
\(\Rightarrow\frac{7}{8}.x=\frac{4}{9}-\left(-\frac{2}{3}\right)\)
\(\Rightarrow\frac{7}{8}.x=\frac{4}{9}+\frac{2}{3}\)
\(\Rightarrow\frac{7}{8}.x=\frac{4}{9}+\frac{6}{9}\)
\(\Rightarrow\frac{7}{8}.x=\frac{10}{9}\)
\(\Rightarrow x=\frac{10}{9}:\frac{7}{8}\)
\(\Rightarrow x=\frac{10}{9}.\frac{8}{7}\)
\(\Rightarrow x=\frac{80}{63}\)
Vây \(x=\frac{80}{63}\)
\(e.\)
\(-\frac{5}{14}:x=-\frac{3}{10}\)
\(\Rightarrow x=-\frac{5}{14}:-\frac{3}{10}\)
\(\Rightarrow x=-\frac{5}{14}.-\frac{10}{3}\)
\(\Rightarrow x=\frac{5}{7}.\frac{5}{3}\)
\(\Rightarrow x=\frac{25}{21}\)
Vậy \(x=\frac{25}{21}\)
\(g.\)
\(\frac{1}{6}+\frac{-5}{7}:x=-\frac{7}{18}\)
\(\Rightarrow-\frac{5}{7}:x=-\frac{7}{18}-\frac{1}{6}\)
\(\Rightarrow-\frac{5}{7}:x=-\frac{7}{18}-\frac{3}{18}\)
\(\Rightarrow-\frac{5}{7}:x=-\frac{10}{18}\)
\(\Rightarrow-\frac{5}{7}:x=-\frac{5}{9}\)
\(\Rightarrow x=-\frac{5}{7}:-\frac{5}{9}\)
\(\Rightarrow x=-\frac{5}{7}.-\frac{9}{5}\)
\(\Rightarrow x=\frac{9}{7}\)
Vậy \(x=\frac{9}{7}\)
a) 4x- 15 = -75 - x
=> 4x + x = -75 + 15
=> 5x = -60
=> x = -60 : 5
=> x = -12
b) 3(x - 7) = 21
=> x - 7 = 21 : 3
=> x - 7 = 7
=> x = 7 + 7
=> x = 14
c) \(-\frac{3}{6}=\frac{x}{-2}=\frac{-18}{y}=\frac{-z}{24}\)
Ta có: +) \(-\frac{3}{6}=\frac{x}{-2}\) => \(\frac{1}{-2}=\frac{x}{-2}\) => x = 1
+) \(\frac{-3}{6}=\frac{-18}{y}\) => \(\frac{-18}{36}=\frac{-18}{y}\) => y = 36
+) \(-\frac{3}{6}=\frac{-z}{24}\) = > \(\frac{-12}{24}=\frac{-z}{24}\) => -z = -12 => z = 12
d) \(-\frac{8}{3}+\frac{-1}{4}< x< \frac{-2}{7}+\frac{-5}{7}\)
=> \(\frac{-32}{12}-\frac{3}{12}< x< \frac{-2-5}{7}\)
=> \(\frac{-35}{12}< x< -1\)
=> x = -2 (x \(\in\)Z)
a) 4x - 15 = -75 - x
4x + x = -75 + 15
5x = -60
=> x = -60 : 5 = -12
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
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}
a, \(\dfrac{7}{18}x-\dfrac{2}{3}=\dfrac{5}{18}\)
\(\dfrac{7}{18}x=\dfrac{5}{18}+\dfrac{2}{3}\)
\(\dfrac{7}{18}x=\dfrac{17}{18}\)
\(x=\dfrac{17}{18}\div\dfrac{7}{18}\)
\(x=\dfrac{17}{7}\)
a) \(\dfrac{7}{18}\).x-\(\dfrac{2}{3}\)=\(\dfrac{5}{18}\)
\(\dfrac{7}{18}\).x =\(\dfrac{5}{18}\)+\(\dfrac{2}{3}\)
\(\dfrac{7}{18}\).x = \(\dfrac{17}{18}\)
x = \(\dfrac{17}{18}\) :\(\dfrac{7}{18}\)
x =\(\dfrac{17}{7}\)
b)\(\dfrac{4}{9}\) - \(\dfrac{7}{8}\).x =\(\dfrac{-2}{3}\)
\(\dfrac{7}{8}\).x =\(\dfrac{4}{9}\)-\(\dfrac{-2}{3}\)
\(\dfrac{7}{8}\).x =\(\dfrac{10}{9}\)
x =\(\dfrac{10}{9}\) : \(\dfrac{7}{8}\)
x =\(\dfrac{80}{63}\)
c)\(\dfrac{1}{6}\)+\(\dfrac{-5}{7}\): \(\dfrac{-7}{18}\)
\(\dfrac{1}{6}\)+\(\dfrac{90}{49}\)
\(\dfrac{589}{294}\)