Tìm x:
a) \(\dfrac{x}{4}=\dfrac{4}{x}\)
b) \(\dfrac{x+7}{15}=-\dfrac{24}{36}\)
c) \(\dfrac{x+1}{8}=\dfrac{2}{x+1}\)
d) \(\dfrac{2x-1}{\left(-3\right)^2}=\dfrac{\left(-3\right)^2}{2x-1}\)
a, đk x khác 0
<=> x^2 = 16 <=> x = 4 ; x = -4 (tm)
b, <=> 36x +252 = -360 <=> x = -17
c. đk x khác -1
<=> (x+1)^2 = 16
TH1 : x + 1 = 4 <=> x = 3 (tm)
TH2 : x + 1 = -4 <=> x = -5 (tm)
d, đk x khác 1/2
<=> (2x-1)^2 = 81
TH1 : 2x - 1 = 9 <=> x = 5 (tm)
TH2 : 2x - 1 = -9 <=> x = -4 (tm)
a: \(\Leftrightarrow x^2=16\)
hay \(x\in\left\{4;-4\right\}\)
b: =>x+7/15=-2/3
=>x+7=-10
hay x=-17
c: \(\Leftrightarrow\left(x+1\right)^2=16\)
\(\Leftrightarrow x+1\in\left\{4;-4\right\}\)
hay \(x\in\left\{3;-5\right\}\)
a) \(\dfrac{x}{4}=\dfrac{4}{x}\)=>x2=4.4=16 =>x2=42
=>x=2 hay x=-2.
b) \(\dfrac{x+7}{15}=-\dfrac{24}{36}\)=>\(\dfrac{x+7}{15}=-\dfrac{2}{3}\)=>x+7=-\(\dfrac{2}{3}.15\)=-10 =>x=-17
c)\(\dfrac{x+1}{8}=\dfrac{2}{x+1}\)=>(x+1)2=2.8=16=42
=>x+1=4 hay x+1=-4
=>x=3 hay x=-5.
d) \(\dfrac{2x-1}{\left(-3\right)^2}=\dfrac{\left(-3\right)^2}{2x-1}\)=>\(\dfrac{2x-1}{9}=\dfrac{9}{2x-1}\)=>(2x-1)2=92
=>2x-1=9 hay 2x-1=-9
=>x=5 hay x=-4.
Tìm số nguyên x, biết:
a) \(\dfrac{-28}{35}=\dfrac{16}{x};\) b) \(\dfrac{x+7}{15}=\dfrac{-24}{36}.\)
\(a.\)
\(\dfrac{-28}{35}=\dfrac{16}{x}\)
\(\Rightarrow x=\dfrac{35\cdot16}{-28}=\dfrac{5\cdot7\cdot4\cdot4}{-7\cdot4}=-20\)
\(b.\)
\(\dfrac{x+7}{15}=\dfrac{-24}{36}\)
\(\Rightarrow x+7=\dfrac{15\cdot-24}{36}=\dfrac{5\cdot3\cdot-12\cdot2}{12\cdot3}=-10\)
\(\Leftrightarrow x=-17\)
\(\dfrac{-2}{9}\)và\(\dfrac{6}{-27}\) b:\(\dfrac{-1}{-5}\)và\(\dfrac{4}{25}\)
Các cặp phân số sau có bằng nhau ko?vì sao?
Bài3: Tìm số nguyên X biết
a)\(\dfrac{-28}{35}\)=\(\dfrac{16}{x}\)
b)\(\dfrac{x+7}{15}\)=\(\dfrac{-24}{36}\)
giúp mình với ae cứu tôi ae cứu tôi :((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((
Bài 2:
a: -2*(-27)=54
6*9=54
=>Hai phân số này bằng nhau
b: -1/-5=1/5=5/25<>4/25
Bài 3:
a: =>16/x=-4/5
=>x=-20
b: =>(x+7)/15=-2/3
=>x+7=-10
=>x=-17
a) \(\dfrac{-2}{9}\) và \(\dfrac{6}{-27}\)
\(\dfrac{6}{-27}=\dfrac{6:\left(-3\right)}{\left(-27\right):\left(-3\right)}=\dfrac{-2}{9}\)
Vậy \(\dfrac{-2}{9}=\dfrac{6}{-27}\)
b) \(\dfrac{-1}{-5}\) và \(\dfrac{4}{25}\)
\(\dfrac{-1}{-5}=\dfrac{\left(-1\right).\left(-5\right)}{\left(-5\right).\left(-5\right)}=\dfrac{5}{25}\)
Do \(5\ne4\Rightarrow\dfrac{5}{25}\ne\dfrac{4}{25}\)
Vậy \(\dfrac{-1}{-5}\ne\dfrac{4}{25}\)
Bài 3
a) \(\dfrac{-28}{35}=\dfrac{16}{x}\)
\(x=\dfrac{35.16}{-28}\)
\(x=-20\)
b) \(\dfrac{x+7}{15}=\dfrac{-24}{36}\)
\(\left(x+7\right).36=15.\left(-24\right)\)
\(36x+252=-360\)
\(36x=-360-252\)
\(36x=-612\)
\(x=\dfrac{-612}{36}\)
\(x=-17\)
Tìm x biết:
a)\(\dfrac{-2}{3}x=\dfrac{4}{15}\) b) \(\dfrac{-7}{19}x=\dfrac{-13}{24}\)
x=\(\dfrac{4}{15}\) : \(\dfrac{-2}{3}\)
x=\(\dfrac{-2}{5}\)
a: Ta có: \(x\cdot\dfrac{-2}{3}=\dfrac{4}{15}\)
\(\Leftrightarrow x=\dfrac{4}{15}:\dfrac{-2}{3}=\dfrac{4}{15}\cdot\dfrac{-3}{2}=\dfrac{-2}{5}\)
b: Ta có: \(x\cdot\dfrac{-7}{19}=\dfrac{-13}{24}\)
\(\Leftrightarrow x=\dfrac{13}{24}:\dfrac{7}{19}=\dfrac{247}{168}\)
\(\dfrac{-28}{35}\)=\(\dfrac{16}{x}\)
\(\dfrac{x+7}{15}\)=\(\dfrac{-24}{36}\)
giúp mình
a: \(\Leftrightarrow\dfrac{16}{x}=\dfrac{-4}{5}\)
hay x=-20
b: =>x+7=-10
hay x=-17
\(-\dfrac{28}{35}=\dfrac{16}{x}\Leftrightarrow-\dfrac{4}{5}=\dfrac{16}{x}\)
\(\Rightarrow-4x=5\cdot16\)
\(\Rightarrow-4x=80\)
\(\Rightarrow x=-20\)
\(\dfrac{x+7}{15}=-\dfrac{24}{36}\Leftrightarrow\)\(\dfrac{x+7}{15}=-\dfrac{2}{3}\)
\(\Rightarrow3\left(x+7\right)=-2\cdot15\)
\(x+7=-30\div3\)
x+7=-10
x=-10-7
x=-17
Tìm x biết: a) \(\dfrac{6}{-x}=\dfrac{x}{-24}\) b) \(x-\dfrac{7}{12}x+\dfrac{3}{8}x=\dfrac{5}{24}\)
c)\(\left(x-\dfrac{1}{3}\right)^2-\dfrac{1}{2}=1\dfrac{3}{4}\) d) \(\dfrac{x-3}{-2}=\dfrac{-8}{x-3}\)
e) \(\dfrac{9}{x}=\dfrac{-35}{105}\) f) \(\left(x-\dfrac{1}{2}\right)\left(-3-\dfrac{x}{2}\right)=0\)
a: =>6/x=x/24
=>x^2=144
=>x=12 hoặc x=-12
b: =>x(1-7/12+3/8)=5/24
=>x*19/24=5/24
=>x=5/24:19/24=5/19
c: =>(x-1/3)^2=1+3/4+1/2=9/4
=>x-1/3=3/2 hoặc x-1/3=-3/2
=>x=11/6 hoặc x=-7/6
d: =>(x-3)^2=16
=>x-3=4 hoặc x-3=-4
=>x=-1 hoặc x=7
e: =>9/x=-1/3
=>x=-27
f: =>x-1/2=0 hoặc -x/2-3=0
=>x=1/2 hoặc x=-6
bài 1: tìm x
(\(\dfrac{5}{6}\) - x + \(\dfrac{7}{12}\)) : (\(\dfrac{11}{24}\) - \(\dfrac{1}{8}\)) = \(\dfrac{11}{36}\)
`(5/6 -x+7/12) : ( 11/24 -1/8)=11/36`
`=>(5/6 -x+7/12) : (11/24 - 3/24)=11/36`
`=>(5/6 -x+7/12) : 8=11/36`
`=>5/6 -x+7/12=11/36 xx 8`
`=>5/6 -x+7/12=22/9`
`=> x+7/12=5/6-22/9`
`=> x+7/12=-29/18`
`=>x=-29/18 -7/12`
`=>x=-79/36`
rút gọn phân thức
a.\(\dfrac{5x-15}{4x+4}:\dfrac{x^2-9}{x^2+2x+1}\)
b.\(\dfrac{6x+48}{7x-7}:\dfrac{x^2-64}{x^2-2x+1}\)
c.\(\dfrac{4x-24}{5x+5}:\dfrac{x^2-36}{x^2+2x+1}\)
d.\(\left(\dfrac{1}{x^2+x}-\dfrac{2-x}{x+1}\right):\left(\dfrac{1}{x}+x-2\right)\)
a.\(\dfrac{5\left(x-3\right)}{4\left(x+1\right)}\) : \(\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x+1\right)^2}\)
= \(\dfrac{5\left(x-3\right)}{4\left(x+1\right)}\). \(\dfrac{\left(x+1\right)^2}{\left(x-3\right)\left(x+3\right)}\)
= \(\dfrac{5\left(x+1\right)}{4\left(x+3\right)}\)
b. \(\dfrac{6\left(x+8\right)}{7\left(x-1\right)}\). \(\dfrac{\left(x-1\right)^2}{\left(x-8\right)\left(x+8\right)}\)
= \(\dfrac{6\left(x-1\right)}{7\left(x-8\right)}\)
c.Tương tự hai câu trên nka!!
d. (\(\dfrac{1}{x\left(x+1\right)}\)-\(\dfrac{2-x}{x+1}\)).(\(\dfrac{x}{x-1}\))
=( \(\dfrac{1}{x\left(x+1\right)}\)-\(\dfrac{2x-x^2}{x\left(x+1\right)}\)). ....
= \(\dfrac{\left(1-x\right)^2}{x\left(x+1\right)}\). ...
= \(\dfrac{x-1}{x+1}\)
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c: \(=\dfrac{4\left(x-6\right)}{5\left(x+1\right)}\cdot\dfrac{\left(x+1\right)^2}{\left(x-6\right)\left(x+6\right)}=\dfrac{4\left(x+1\right)}{5\left(x+6\right)}\)
a: \(=\dfrac{5\left(x-3\right)}{4\left(x+1\right)}\cdot\dfrac{\left(x+1\right)^2}{\left(x-3\right)\left(x+3\right)}=\dfrac{5\left(x+1\right)}{4\left(x+3\right)}\)
b: \(=\dfrac{6\left(x+8\right)}{7\left(x-1\right)}\cdot\dfrac{\left(x-1\right)^2}{\left(x-8\right)\left(x+8\right)}=\dfrac{6\left(x-1\right)}{7\left(x-8\right)}\)
Viết tập hợp A các số tự nhiên x sao cho:
a, \(\dfrac{-36}{9}\) ≤ x < \(\dfrac{-15}{5}\)
b, \(\dfrac{-27}{3}\) < x ≤ \(\dfrac{12}{4}\)
c, \(\dfrac{-21}{7}\) < x < \(\dfrac{-12}{6}\)
a: \(\Leftrightarrow-4< =x< =-3\)
hay \(x\in\varnothing\)
b: =>-9<x<=3
hay \(x\in\left\{0;1;2;3\right\}\)
