Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1. a, \(\dfrac{x}{7}=\dfrac{9}{y}\Leftrightarrow xy=9.7\)
<=> xy = 63
=> x; y \(\inƯ\left(63\right)\)
Lại có x > y nên ta có bảng :
| x | 63 | -1 | 21 | -3 | 9 | -7 |
| y | 1 | -63 | 3 | -21 | 7 | -9 |
@Đặng Hoài An
1. b, \(\dfrac{-2}{x}=\dfrac{y}{5}\Leftrightarrow-2.5=xy\)
<=> -10 = xy
=> x; y \(\inƯ\left(10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)
Lại có : x < 0 < y
=> x = -1; -2; -5; -10
Tương ứng y = 10; 5; 2; 1
@Đặng Hoài An
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
Bài 1:
a: =>3x-3-4=0
=>3x=7
hay x=7/3
b: =>2x-2+3x+6=0
=>5x+4=0
hay x=-4/5
c: =>\(4x^2+4x-1=0\)
hay \(x\in\left\{\dfrac{-1+\sqrt{2}}{2};\dfrac{-1-\sqrt{2}}{2}\right\}\)
d: \(\Leftrightarrow3x-3+2x-4+6=0\)
=>5x+1=0
hay x=-1/5
2. Tính:
a, \(\dfrac{-1}{20}+\dfrac{-1}{30}+\dfrac{-1}{42}+\dfrac{-1}{56}+\dfrac{-1}{72}+\dfrac{-1}{90}\)
=\(\left(\dfrac{-1}{20}+\dfrac{-1}{72}\right)+\left(\dfrac{-1}{30}+\dfrac{-1}{90}\right)+\left(\dfrac{-1}{42}+\dfrac{-1}{56}\right)\)
=\(\left(\dfrac{-18}{360}+\dfrac{-5}{360}\right)+\left(\dfrac{-3}{90}+\dfrac{-1}{90}\right)+\left(\dfrac{-4}{168}+\dfrac{-3}{168}\right)\)
=\(\dfrac{-23}{360}+\dfrac{-4}{90}+\dfrac{-7}{168}\)
=\(\dfrac{-23}{360}+\dfrac{-16}{360}+\dfrac{-15}{360}\)=\(\dfrac{-54}{360}=\dfrac{-3}{20}\)
b, \(\dfrac{5}{2.1}+\dfrac{4}{1.11}+\dfrac{3}{11.2}+\dfrac{1}{2.15}+\dfrac{13}{15.4}\)
=\(\dfrac{5}{2}+\dfrac{4}{1}.\dfrac{1}{11}+\dfrac{1}{11}.\dfrac{3}{2}+\dfrac{1}{2}.\dfrac{1}{15}+\dfrac{1}{15}.\dfrac{13}{4}\)
=\(\dfrac{5}{2}+\dfrac{1}{11}.\left(\dfrac{4}{1}+\dfrac{3}{2}\right)+\dfrac{1}{15}.\left(\dfrac{1}{2}+\dfrac{13}{4}\right)\)
=\(\dfrac{5}{2}+\dfrac{1}{11}.\dfrac{11}{2}+\dfrac{1}{15}.\dfrac{15}{4}\)
=\(\dfrac{5}{2}+\dfrac{1}{2}+\dfrac{1}{4}\)
=\(\dfrac{10}{4}+\dfrac{2}{4}+\dfrac{1}{4}\)
=\(\dfrac{13}{4}\)
3. Tìm x
a, \(\dfrac{x-5}{8}=\dfrac{18}{x-5}\)
\(\left(x-5\right).\left(x-5\right)=8.18\)
\(\left(x-5\right)^2=144\)
\(x-5=\sqrt{144}\)
\(x-5=12\)
\(x=12+5\)
\(x=17\)
b,\(\left(x-2\right)^{10}=\left(2-x\right)^8\)
\(x^{10}-2^{10}=x^8-2^8\)
\(x^{10}+x^8=2^{10}+2^8\)
\(\Rightarrow x=2\)
\(4)\)
\(\dfrac{-\left(-x\right)}{5}-\dfrac{2}{10}=\dfrac{1}{-5}-\dfrac{7}{50}\)
\(\Leftrightarrow\dfrac{x}{5}-\dfrac{2}{10}=\dfrac{1}{-5}-\dfrac{7}{50}\)
\(\dfrac{2x}{10}-\dfrac{2}{10}=\dfrac{-10}{50}-\dfrac{7}{50}\)
\(\Leftrightarrow\dfrac{2x-2}{10}=\dfrac{-10-7}{50}\)
\(\dfrac{2x-2}{10}=\dfrac{-17}{50}\)
\(\Leftrightarrow50\left(2x-2\right)=-17.10\)
\(100x-100=-170\)
\(100x=-170+100=-70\)
\(x=-70:100=\dfrac{-7}{10}\)
\(\dfrac{x+1}{5}=\dfrac{7}{x-1}\)
\(\left(x+1\right)\left(x-1\right)5.7\)
\(x\left(x-1\right)+1\left(x-1\right)=35\)
\(x^2-x+x-1=35\)
\(x^2-1=35\)
\(x^2=36\)
\(\Leftrightarrow x=\left\{\pm6\right\}\)
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 4:
Gọi số cần tìm là x
Theo đề, ta có:
\(\dfrac{x+19}{x+17}=\dfrac{3}{5}\)
=>5x+95=3x+51
=>2x=-44
hay x=-22
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)
\(3\cdot\left(3x-\dfrac{1}{2}\right)^3+\dfrac{1}{9}=0\\ 3\cdot\left(3x-\dfrac{1}{2}\right)^3=0-\dfrac{1}{9}\\ 3\cdot\left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{9}\\ \left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{9}:3\\ \left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{27}\\ \left(3x-\dfrac{1}{2}\right)^3=\left(\dfrac{-1}{3}\right)^3\\ \Leftrightarrow3x-\dfrac{1}{2}=\dfrac{-1}{3}\\ 3x=\dfrac{-1}{3}+\dfrac{1}{2}\\ 3x=\dfrac{1}{6}\\ x=\dfrac{1}{6}:3\\ x=\dfrac{1}{18}\)
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