\(\dfrac{36}{x^6}\)-\(\dfrac{24}{x^3}\)+4
Tìm x: a)\(\dfrac{1}{-x}=-4\) b) \(\dfrac{x+7}{15}\dfrac{-24}{36}\) c)\(\dfrac{x}{-3}=\dfrac{-12}{x}\) d)\(\dfrac{-2}{x-3}=\dfrac{x-3}{-8}\)
a: =>1/x=4
hay x=1/4
b: =>x+7=-10
=>x=-17
c: =>x2=36
=>x=6 hoặc x=-6
d: =>(x-3)2=16
=>x-3=4 hoặc x-3=-4
=>x=7 hoặc x=-1
\(a,\dfrac{1}{-x}=-4\\ \Rightarrow\left(-x\right)\left(-4\right)=1\\ \Rightarrow4x=1\\ \Rightarrow x=\dfrac{1}{4}\\ b,\dfrac{x+7}{15}=\dfrac{-24}{36}\\ \Rightarrow\dfrac{x+7}{15}=\dfrac{-2}{3}\\ \Rightarrow3\left(x+7\right)=-2.15\\ \Rightarrow3x+21=-30\\ \Rightarrow3x=-51\\ \Rightarrow x=-17\)
\(c,\dfrac{x}{-3}=\dfrac{-12}{x}\\ \Rightarrow x.x=\left(-3\right)\left(-12\right)\\ \Rightarrow x^2=36\\ \Rightarrow x=\pm6\)
\(d,\dfrac{-2}{x-3}=\dfrac{x-3}{-8}\\ \Rightarrow\left(x-3\right)\left(x-3\right)=\left(-2\right)\left(-8\right)\\ \Rightarrow\left(x-3\right)^2=16\\ \Rightarrow\left[{}\begin{matrix}x-3=4\\x-3=-4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=7\\x=-1\end{matrix}\right.\)
a.\(\Leftrightarrow4x=1\Leftrightarrow x=\dfrac{1}{4}\)
b.\(\Leftrightarrow\left(x+7\right).36=-360\Leftrightarrow x+7=-10\Rightarrow x=-17\)
c.\(x^2=36\Leftrightarrow x=6;-6\)
d.\(-2.-8=\left(x-3\right)\left(x-3\right)\Leftrightarrow16=x^2-6x+9\Leftrightarrow x^2-6x-7=0\Rightarrow x=7;-1\)
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.
thực hiện phép tính
\(\dfrac{x}{x-3}-\dfrac{6}{x}-\dfrac{9}{x^2-3x}\)
\(\dfrac{7}{x}-\dfrac{x}{x+6}+\dfrac{36}{x^2+6x}\)
\(\dfrac{6}{x-3}-\dfrac{2x-16}{x^2-9}-\dfrac{4}{x+3}\)
a) Ta có: \(\dfrac{x}{x-3}-\dfrac{6}{x}-\dfrac{9}{x^2-3x}\)
\(=\dfrac{x^2}{x\left(x-3\right)}-\dfrac{6\left(x-3\right)}{x\left(x-3\right)}-\dfrac{9}{x\left(x-3\right)}\)
\(=\dfrac{x^2-6x+18-9}{x\left(x-3\right)}\)
\(=\dfrac{\left(x-3\right)^2}{x\left(x-3\right)}=\dfrac{x-3}{x}\)
b) Ta có: \(\dfrac{7}{x}-\dfrac{x}{x+6}+\dfrac{36}{x^2+6x}\)
\(=\dfrac{7\left(x+6\right)-x^2+36}{x\left(x+6\right)}\)
\(=\dfrac{7x+42-x^2+36}{x\left(x+6\right)}\)
\(=\dfrac{-\left(x^2-7x-78\right)}{x\left(x+6\right)}\)
\(=\dfrac{-\left(x^2-13x+6x-78\right)}{x\left(x+6\right)}\)
\(=\dfrac{-\left[x\left(x-13\right)+6\left(x-13\right)\right]}{x\left(x+6\right)}\)
\(=\dfrac{13-x}{x}\)
c) Ta có: \(\dfrac{6}{x-3}-\dfrac{2x-6}{x^2-9}-\dfrac{4}{x+3}\)
\(=\dfrac{6\left(x+3\right)-2x+6-4\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{6x+18-2x+6-4x+12}{\left(x-3\right)\left(x+3\right)}=\dfrac{36}{\left(x-3\right)\left(x+3\right)}\)
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`
\(\dfrac{-2}{9}\)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 các số nguyên x,y biết:
a)\(\dfrac{6}{2x+1}=\dfrac{2}{7}\)
b) \(\dfrac{24}{7x-3}=\dfrac{-4}{25}\)
c) \(\dfrac{4}{x-6}=\dfrac{y}{24}=\dfrac{-12}{18}\)
d) \(\dfrac{-1}{5}\le\dfrac{x}{8}\le\dfrac{1}{4}\)
e) \(\dfrac{x+46}{20}=x\dfrac{2}{5}\)
f) \(y\dfrac{5}{y}=\dfrac{86}{y}\) ( \(x\dfrac{2}{5};y\dfrac{5}{y}\) là các hỗn số)
a,\(\dfrac{6}{2x+1}=\dfrac{2}{7}\)
⇒\(\dfrac{6}{2x+1}=\dfrac{6}{21}\)
⇒\(2x+1=21\)
\(2x=21-1\)
\(2x=20\)
⇒\(x=10\)
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
Câu 1)
1) \(\dfrac{11}{24}\)−\(\dfrac{5}{41}\)+\(\dfrac{13}{24}\)+0,5−\(\dfrac{36}{41}\)=
2)12÷\(\left(\dfrac{3}{4}-\dfrac{5}{6}\right)^2\)=
3) (\(1+\dfrac{2}{3}-\dfrac{1}{4}\))\(\left(0,8-\dfrac{3}{4}\right)^2\) =
4)\(16\dfrac{2}{7}\)÷(\(\dfrac{-3}{5}\))+\(28\dfrac{2}{7}\)÷\(\dfrac{3}{5}\)
5)\(\left(2^2\div\dfrac{4}{3}-\dfrac{1}{2}\right)\times\dfrac{6}{5}-17\)
6)\(\left(\dfrac{1}{3}\right)^{50}\times\left(-9\right)^{25}-\dfrac{2}{3}\div4\)
1: \(\dfrac{11}{24}-\dfrac{5}{41}+\dfrac{13}{24}+0,5-\dfrac{36}{41}\)
\(=\left(\dfrac{11}{24}+\dfrac{13}{24}\right)-\left(\dfrac{5}{41}+\dfrac{36}{41}\right)+\dfrac{1}{2}\)
\(=1-1+\dfrac{1}{2}=\dfrac{1}{2}\)
2: \(12:\left(\dfrac{3}{4}-\dfrac{5}{6}\right)^2\)
\(=12:\left(\dfrac{9}{12}-\dfrac{10}{12}\right)^2\)
\(=12:\left(-\dfrac{1}{12}\right)^2=12:\dfrac{1}{144}=12\cdot144=1368\)
3: \(\left(1+\dfrac{2}{3}-\dfrac{1}{4}\right)\cdot\left(0,8-\dfrac{3}{4}\right)^2\)
\(=\dfrac{12+8-3}{12}\cdot\left(\dfrac{4}{5}-\dfrac{3}{4}\right)^2\)
\(=\dfrac{17}{12}\cdot\left(\dfrac{16-15}{20}\right)^2\)
\(=\dfrac{17}{12}\cdot\dfrac{1}{400}=\dfrac{17}{4800}\)
4: \(16\dfrac{2}{7}:\left(-\dfrac{3}{5}\right)+28\dfrac{2}{7}:\dfrac{3}{5}\)
\(=\dfrac{5}{3}\cdot\left(-16-\dfrac{2}{7}\right)+\dfrac{5}{3}\cdot\left(28+\dfrac{2}{7}\right)\)
\(=\dfrac{5}{3}\left(-16-\dfrac{2}{7}+28+\dfrac{2}{7}\right)\)
\(=12\cdot\dfrac{5}{3}=20\)
5: \(\left(2^2:\dfrac{4}{3}-\dfrac{1}{2}\right)\cdot\dfrac{6}{5}-17\)
\(=\left(4\cdot\dfrac{3}{4}-\dfrac{1}{2}\right)\cdot\dfrac{6}{5}-17\)
\(=\dfrac{5}{2}\cdot\dfrac{6}{5}-17=3-17=-14\)
6: \(\left(\dfrac{1}{3}\right)^{50}\cdot\left(-9\right)^{25}-\dfrac{2}{3}:4\)
\(=\left(\dfrac{1}{3}\right)^{50}\cdot\left(-1\right)\cdot3^{50}-\dfrac{2}{3\cdot4}\)
\(=-1-\dfrac{2}{12}=-1-\dfrac{1}{6}=-\dfrac{7}{6}\)
Cho hai biểu thức:
A = \(\dfrac{24}{\sqrt{x}+6}\) và B = \(\dfrac{\sqrt{x}}{\sqrt{x}+6}+\dfrac{1}{\sqrt{x}-6}+\dfrac{17\sqrt{x}+30}{x-36}\) với \(x\ge0;x\ne36\)
c) Biểu thức B sau khi thu gọn được B = \(\dfrac{\sqrt{x}+6}{\sqrt{x}-6}\). Tìm các giá trị của x để AB \(\le12\)
Ta có :
\(A.B=\dfrac{24}{\sqrt{x}+6}.\dfrac{\sqrt{x}+6}{\sqrt{x}-6}\)
\(=\dfrac{24}{\sqrt{x}-6}\)
Để \(AB\le12\Leftrightarrow\dfrac{24}{\sqrt{x}-6}\le12\)
\(\Leftrightarrow\dfrac{24-12\left(\sqrt{x}-6\right)}{\sqrt{x}-6}\le0\)
\(\Leftrightarrow24-12\sqrt{x}+72\le0\)
\(\Leftrightarrow-12\sqrt{x}\le-96\)
\(\Leftrightarrow\sqrt{x}\ge8\)
\(\Leftrightarrow x\ge64\)
Vậy \(x\ge64\) thì \(AB\le12\)