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Ta có: 1/3 + −2/5+ 1/6 + −1/5 ≤ x < −3/4+2/7+-1/4+3/5+5/7
⇒10-12+5-6/30≤ x< -105+40-35+84+100/140
⇒-3/30≤ x <84/140
⇒-0,1≤ x < 0,6
⇒x=0
bài 3:
a, đặt \(\dfrac{x}{12}=\dfrac{y}{9}=\dfrac{z}{5}=k\)
=>x=12k,y=9k,z=5k
ta có: ayz=20=> 12k.9k.5k=20
=> (12.9.5)k^3=20
=>540.k^3=20
=>k^3=20/540=1/27
=>k=1/3
=>x=12.1/3=4
y=9.1/3=3
z=5.1/3=5/3
vậy x=4,y=3,z=5/3
b,ta có: \(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{3}=\dfrac{x^2}{25}=\dfrac{y^2}{49}=\dfrac{z^2}{9}\)
A/D tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{3}=\dfrac{x^2}{25}=\dfrac{y^2}{49}=\dfrac{z^2}{9}=\dfrac{x^2+y^2-z^2}{25+49-9}=\dfrac{585}{65}=9\)
=>x=5.9=45
y=7.9=63
z=3*9=27
vậy x=45,y=63,z=27
\(4)\)
\(\dfrac{-\left(-x\right)}{5}-\dfrac{2}{10}=\dfrac{1}{-5}-\dfrac{7}{50}\)
\(\Leftrightarrow\dfrac{x}{5}-\dfrac{2}{10}=\dfrac{1}{-5}-\dfrac{7}{50}\)
\(\dfrac{2x}{10}-\dfrac{2}{10}=\dfrac{-10}{50}-\dfrac{7}{50}\)
\(\Leftrightarrow\dfrac{2x-2}{10}=\dfrac{-10-7}{50}\)
\(\dfrac{2x-2}{10}=\dfrac{-17}{50}\)
\(\Leftrightarrow50\left(2x-2\right)=-17.10\)
\(100x-100=-170\)
\(100x=-170+100=-70\)
\(x=-70:100=\dfrac{-7}{10}\)
\(\dfrac{x+1}{5}=\dfrac{7}{x-1}\)
\(\left(x+1\right)\left(x-1\right)5.7\)
\(x\left(x-1\right)+1\left(x-1\right)=35\)
\(x^2-x+x-1=35\)
\(x^2-1=35\)
\(x^2=36\)
\(\Leftrightarrow x=\left\{\pm6\right\}\)
bạn có thể giải đc các bài còn lại k ? K phải mk ép bạn đâu nhưng nếu bạn lm đc thì giúp mk nha
bài 1:
a) \(4\dfrac{1}{2}x:\dfrac{5}{12}=0,5\) ; b)\(1,5+1\dfrac{1}{4}x=\dfrac{2}{3}\)
\(\dfrac{9}{2}x:\dfrac{5}{12}=\dfrac{1}{2}\) \(\dfrac{3}{2}+\dfrac{5}{4}x=\dfrac{2}{3}\)
\(\dfrac{9}{2}x\) \(=\dfrac{1}{2}.\dfrac{5}{12}\) \(\dfrac{5}{4}x=\dfrac{2}{3}-\dfrac{3}{2}\)
\(\dfrac{9}{2}x\) \(=\dfrac{5}{24}\) \(\dfrac{5}{4}x=\dfrac{-5}{6}\)
\(x\) \(=\dfrac{5}{24}:\dfrac{9}{2}\) \(x=\dfrac{-5}{6}:\dfrac{5}{4}\)
\(x\) \(=\dfrac{5}{108}\) \(x=\dfrac{-2}{3}\)
c) Cho mình hỏi x ở đâu vậy ???
d)\(\left(x-5\right):\dfrac{1}{3}=\dfrac{2}{5}\) e)\(\left(4,5-2x\right):\dfrac{3}{4}=1\dfrac{1}{3}\)
\(\left(x-5\right)\) \(=\dfrac{2}{5}.\dfrac{1}{3}\) \(\left(\dfrac{9}{2}-2x\right):\dfrac{3}{4}=\dfrac{4}{3}\)
\(x-5\) \(=\dfrac{2}{15}\) \(\dfrac{9}{2}-2x\) =\(\dfrac{4}{3}.\dfrac{3}{4}\)
\(x\) \(=\dfrac{2}{15}+5\) \(\dfrac{9}{2}-2x=1\)
\(x\) \(=\dfrac{77}{15}\) \(2x=\dfrac{9}{2}-1\)
f) \(\left(2,7x-1\dfrac{1}{2}x\right):\dfrac{2}{7}=\dfrac{-21}{7}\) \(2x=\dfrac{7}{2}\)
\(\left(\dfrac{27}{10}x-\dfrac{3}{2}x\right):\dfrac{2}{7}=-3\) \(x=\dfrac{7}{2}:2\)
\(\left[x\left(\dfrac{27}{10}-\dfrac{3}{2}\right)\right]=-3.\dfrac{2}{7}\) \(x=\dfrac{7}{4}\)
\(x.\dfrac{6}{5}=\dfrac{-6}{7}\)
\(x=\dfrac{-6}{7}:\dfrac{6}{5}\)
\(x=\dfrac{-5}{7}\)
bài 2:
Theo bài ra ta có :\(\dfrac{a}{27}=\dfrac{-5}{9}=\dfrac{-45}{b}\)
\(\Rightarrow9a=27.\left(-5\right)\Rightarrow a=\dfrac{27.\left(-5\right)}{9}=-15\)
\(\Rightarrow\left(-5\right)b=\left(-45\right).9\Rightarrow b=\dfrac{\left(-45\right).9}{-5}=81\)
Vậy \(a=-15;b=81\)
\(.2.\)
\(a.\)
\(2x+\dfrac{1}{2}=-\dfrac{5}{3}\)
\(\Rightarrow2x=-\dfrac{5}{3}-\dfrac{1}{2}=-\dfrac{13}{6}\)
\(\Rightarrow x=-\dfrac{13}{6}:2=-\dfrac{13}{12}\)
Vậy : \(x=-\dfrac{13}{12}\)
\(b.\)
\(\dfrac{1}{7}-\dfrac{3}{5}x=\dfrac{3}{5}\)
\(\Rightarrow\dfrac{3}{5}x=\dfrac{1}{7}-\dfrac{3}{5}=-\dfrac{16}{35}\)
\(\Rightarrow x=-\dfrac{16}{35}:\dfrac{3}{5}=-\dfrac{16}{21}\)
Vậy : \(x=-\dfrac{16}{21}\)
\(c.\)
\(\dfrac{3}{4}x+\dfrac{1}{2}=-\dfrac{3}{5}\)
\(\Rightarrow\dfrac{3}{4}x=-\dfrac{3}{5}-\dfrac{1}{2}=-\dfrac{11}{10}\)
\(\Rightarrow x=-\dfrac{11}{10}:\dfrac{3}{4}=-\dfrac{22}{15}\)
Vậy : \(x=-\dfrac{22}{15}\)
\(d.\)
\(-\dfrac{2}{15}-x=-\dfrac{3}{10}\)
\(\Rightarrow x=-\dfrac{2}{15}-\left(-\dfrac{3}{10}\right)=\dfrac{1}{6}\)
Vậy : \(x=\dfrac{1}{6}\)
Bài 2.
A = -3/5 + ( -2/5 + 2 )
A = -3/5 + ( -2/5 + 10/5 )
A = -3/5 + 8/5
A = 5/5
A = 1
--------------------------------------------------------
B = 3/7 + ( -1/5 + -3/7 )
B = 3/7 + ( -7/35 + -15/35 )
B = 3/7 + ( -22/35 )
B = 15/35 + ( -22/35 )
B = -1/5
-----------------------------------------------------
C = ( -5/24 + 0,75 + 7/12 ) : ( -2 . 1/8 )
C = ( -5/24 + 3/4 + 7/12 ) : ( -1/4 )
C = 9/8 : ( -1/4 )
C = 9/8 . ( -4 )
C = -9/2
Bài 3 .
a) 4/7 - x = 1/2 . x + 2/7
<=> -x - x = 1/2 - 4/7 + 2/7
<=> -2x = 3/14
<=> x = 3/14 . ( -1/2 )
<=> x = -3/28
Vậy x = -3/28
b) x : 3 1/5 = 1 1/2
<=> x : 16/5 = 3/2
<=> x = 3/2 . 16/5
<=> x = 24/5
Vậy x = 24/5
c) x . 3/4 = -1 5/8
<=> x . 3/4 = -13/8
<=> x = -13/8 . 4/3
<=> x = -13/6
Vậy x = -13/6
Bài 1:
a)
\(\dfrac{x-1}{9}=\dfrac{8}{3}\\ \Leftrightarrow\dfrac{x-1}{9}=\dfrac{24}{9}\\ \Leftrightarrow x-1=24\\ x=24+1\\ x=25\)
b)
\(\left(\dfrac{3x}{7}+1\right):\left(-4\right)=\dfrac{-1}{8}\\ \dfrac{3x}{7}+1=\dfrac{-1}{8}\cdot\left(-4\right)\\ \dfrac{3x}{7}+1=\dfrac{1}{2}\\ \dfrac{3x}{7}=\dfrac{1}{2}-1\\ \dfrac{3x}{7}=\dfrac{-1}{2}\\ 3x=\dfrac{-1}{2}\cdot7\\ 3x=\dfrac{-7}{2}\\ x=\dfrac{-7}{2}:3\\ x=\dfrac{-7}{6}\)
c)
\(x+\dfrac{7}{12}=\dfrac{17}{18}-\dfrac{1}{9}\\ x+\dfrac{7}{12}=\dfrac{5}{6}\\ x=\dfrac{5}{6}-\dfrac{7}{12}\\ x=\dfrac{1}{4}\)
d)
\(0,5x-\dfrac{2}{3}x=\dfrac{7}{12}\\ \dfrac{1}{2}x-\dfrac{2}{3}x=\dfrac{7}{12}\\ x\cdot\left(\dfrac{1}{2}-\dfrac{2}{3}\right)=\dfrac{7}{12}\\ \dfrac{-1}{6}x=\dfrac{7}{12}\\ x=\dfrac{7}{12}:\dfrac{-1}{6}\\ x=\dfrac{-7}{2}\)
e)
\(\dfrac{29}{30}-\left(\dfrac{13}{23}+x\right)=\dfrac{7}{46}\\ \dfrac{29}{30}-\dfrac{13}{23}-x=\dfrac{7}{46}\\ \dfrac{277}{690}-x=\dfrac{7}{46}\\ x=\dfrac{277}{690}-\dfrac{7}{46}\\ x=\dfrac{86}{345}\)
f)
\(\left(x+\dfrac{1}{4}-\dfrac{1}{3}\right):\left(2+\dfrac{1}{6}-\dfrac{1}{4}\right)=\dfrac{7}{46}\\ \left(x-\dfrac{1}{12}\right):\dfrac{23}{12}=\dfrac{7}{46}\\ x-\dfrac{1}{12}=\dfrac{7}{46}\cdot\dfrac{23}{12}\\ x-\dfrac{1}{12}=\dfrac{7}{24}\\ x=\dfrac{7}{24}+\dfrac{1}{12}\\ x=\dfrac{3}{8}\)
g)
\(\dfrac{13}{15}-\left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{7}{10}\\ \left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{13}{15}-\dfrac{7}{10}\\ \left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{1}{6}\\ \dfrac{13}{21}+x=\dfrac{1}{6}:\dfrac{7}{12}\\ \dfrac{13}{21}+x=\dfrac{2}{7}\\ x=\dfrac{2}{7}-\dfrac{13}{21}\\ x=\dfrac{-1}{3}\)
h)
\(2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|-\dfrac{3}{2}=\dfrac{1}{4}\\ 2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{1}{4}+\dfrac{3}{2}\\ 2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{4}\\ \left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{4}:2\\ \left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{8}\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{8}\\\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{-7}{8}\end{matrix}\right.\\ \dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{8}\\ \dfrac{1}{2}x=\dfrac{7}{8}+\dfrac{1}{3}\\ \dfrac{1}{2}x=\dfrac{29}{24}\\ x=\dfrac{29}{24}:\dfrac{1}{2}\\ x=\dfrac{29}{12}\\ \dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{-7}{8}\\ \dfrac{1}{2}x=\dfrac{-7}{8}+\dfrac{1}{3}\\ \dfrac{1}{2}x=\dfrac{-13}{24}\\ x=\dfrac{-13}{24}:\dfrac{1}{2}\\ x=\dfrac{-13}{12}\)
i)
\(3\cdot\left(3x-\dfrac{1}{2}\right)^3+\dfrac{1}{9}=0\\ 3\cdot\left(3x-\dfrac{1}{2}\right)^3=0-\dfrac{1}{9}\\ 3\cdot\left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{9}\\ \left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{9}:3\\ \left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{27}\\ \left(3x-\dfrac{1}{2}\right)^3=\left(\dfrac{-1}{3}\right)^3\\ \Leftrightarrow3x-\dfrac{1}{2}=\dfrac{-1}{3}\\ 3x=\dfrac{-1}{3}+\dfrac{1}{2}\\ 3x=\dfrac{1}{6}\\ x=\dfrac{1}{6}:3\\ x=\dfrac{1}{18}\)
a: \(\Leftrightarrow x\cdot\dfrac{62}{7}=\dfrac{29}{9}\cdot\dfrac{56}{3}=\dfrac{1624}{27}\)
hay \(x=\dfrac{1624}{27}:\dfrac{62}{7}=\dfrac{5684}{837}\)
b: \(\Leftrightarrow\dfrac{1}{5}:x=\dfrac{12}{35}\)
nên \(x=\dfrac{1}{5}:\dfrac{12}{35}=\dfrac{1}{5}\cdot\dfrac{35}{12}=\dfrac{7}{12}\)
c: \(\Leftrightarrow\left|2x+\dfrac{1}{3}\right|=\dfrac{30-7}{42}=\dfrac{23}{42}\)
=>2x+1/3=23/42 hoặc 2x+1/3=-23/42
=>2x=3/14 hoặc 2x=-37/42
=>x=3/28 hoặc x=-37/84
BÀi 1
Để A \(\in\) Z
=>\(\left(n+2\right)⋮\left(n-5\right)\)
=>\([\left(n-5\right)+7]⋮\left(n-5\right)\)
=>\(7⋮\left(n-5\right)\)
=>\(n-5\in\left\{1;7;-1;-7\right\}\)
=>\(n\in\left\{6;13;4;-2\right\}\)
Vậy \(n\in\left\{6;13;4;-2\right\}\)
Câu 2
(xn)m=xm.n
2:
(xn)m = xn . m