Khi thêm số tự nhiên m, phân số trở thành \(\dfrac{43+m}{60+m}\)

Theo đề bài, ta có:

\(\dfrac{m+43}{m+60}=\dfrac{3}{4}\)

\(\Leftrightarrow4\times\left(m+43\right)=3\times\left(m+60\right)\)

\(\Leftrightarrow4m+172=3m+180\\ \Leftrightarrow m=8\)

Đáp số: m = 8

Giải được rùi nè

Bằng 8

a) \(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{7};x+y+z=56\)

\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{x+y+z}{2+5+7}=\dfrac{56}{14}=4\)

\(\Rightarrow\left\{{}\begin{matrix}x=4.2=8\\y=4.5=20\\z=4.7=28\end{matrix}\right.\)

b) \(\dfrac{x}{1,1}=\dfrac{y}{1,3}=\dfrac{z}{1,4}\left(1\right);2x-y=5,5\)

\(\left(1\right)\Rightarrow\dfrac{2x-y}{1,1.2-1,3}=\dfrac{5,5}{0,9}\)

\(\Rightarrow\left\{{}\begin{matrix}x=1,1.\dfrac{5,5}{0,9}=\dfrac{6,05}{0,9}\\y=1,3.\dfrac{5,5}{0,9}=\dfrac{7,15}{0,9}\\z=\dfrac{1,4}{1,1}.x=\dfrac{1,4}{1,1}.\dfrac{6,05}{0,9}=\dfrac{8,47}{0,99}\end{matrix}\right.\)

d) \(\dfrac{x}{2}=\dfrac{x}{3}=\dfrac{z}{5};xyz=-30\)

\(\dfrac{x}{2}=\dfrac{x}{3}=\dfrac{z}{5}=\dfrac{xyz}{2.3.5}=\dfrac{-30}{30}=-1\)

\(\Rightarrow\left\{{}\begin{matrix}x=2.\left(-1\right)=-2\\y=3.\left(-1\right)=-3\\z=5.\left(-1\right)=-5\end{matrix}\right.\)

a) �2=�5=�7;�+�+�=562x​=5y​=7z​;x+y+z=56

�2=�5=�7=�+�+�2+5+7=5614=42x​=5y​=7z​=2+5+7x+y+z​=1456​=4

⇒{�=4.2=8�=4.5=20�=4.7=28⇒⎩⎨⎧​x=4.2=8y=4.5=20z=4.7=28​

b) �1,1=�1,3=�1,4(1);2�−�=5,51,1x​=1,3y​=1,4z​(1);2x−y=5,5

(1)⇒2�−�1,1.2−1,3=5,50,9(1)⇒1,1.2−1,32x−y​=0,95,5​
    

⇒⎩⎨⎧​x=1,1.0,95,5​=0,96,05​y=1,3.0,95,5​=0,97,15​z=1,11,4​.x=1,11,4​.0,96,05​=0,998,47​​

d) �2=�3=�5;���=−302x​=3x​=5z​;xyz=−30

�2=�3=�5=���2.3.5=−3030=−12x​=3x​=5z​=2.3.5xyz​=30−30​=−1

⇒{�=2.(−1)=−2�=3.(−1)=−3�=5.(−1)=−5⇒⎩⎨⎧​x=2.(−1)=−2y=3.(−1)=−3z=5.(−1)=−5​
 

a) 9x-1/4=3/2

=>9x=3/2+1/4

=>9x=7/4

=>x=7/4:9

=>x=7/36

Vậy x=7/36

b)(4x+2):2,5=3,2:0,5

=>(4x+2):2,5=6,4

=>4x+2=6,4.2,5

=>4x+2=16

=>4x=16-2

=>4x=14

=>x=14:4

=>x=7/2

Vậy x=7/2

c) 5,4/x-2=6/7

=>5,4/x=6/7+2

=>5,4/x=20/7

=>x=5,4 :20/7

=>x=1,89

Vậy x= 1,89

d) 0,5:2=3:(2x+7)

=>3:(2x+7)=0,25

=>2x+7=3:0,25

=>2x+7=12

=>2x=12-7

=>2x=5

=>x=5/2

Vậy x=5/2

a) 9x-1/4=3/2

=>9x=3/2+1/4

=>9x=7/4

=>x=7/4:9

=>x=7/36

Vậy x=7/36

b)(4x+2):2,5=3,2:0,5

=>(4x+2):2,5=6,4

=>4x+2=6,4.2,5

=>4x+2=16

=>4x=16-2

=>4x=14

=>x=14:4

=>x=7/2

Vậy x=7/2

c) 5,4/x-2=6/7

=>5,4/x=6/7+2

=>5,4/x=20/7

=>x=5,4 :20/7

=>x=1,89

Vậy x= 1,89

d) 0,5:2=3:(2x+7)

=>3:(2x+7)=0,25

=>2x+7=3:0,25

=>2x+7=12

=>2x=12-7

=>2x=5

=>x=5/2

Vậy x=5/2

\(...=\dfrac{5}{2}+\dfrac{6}{9}=\dfrac{5}{2}+\dfrac{2}{3}=\dfrac{15}{6}+\dfrac{4}{6}=\dfrac{19}{6}\)

40 chiếc xe đạp ứng với phân số là:

\(1-\dfrac{3}{5}-\left(1-\dfrac{1}{3}\right)\times\dfrac{2}{7}=\dfrac{2}{7}\)(số xe đạp)

Tống số xe đạp là:

\(40:\dfrac{2}{7}=140\)(chiếc xe)

đ/s:..

a) \(...=16.\left(36-37\right)=16.\left(-1\right)=-16\)

b) \(...=9.3+9.50=9.\left(3+50\right)=9.53=477\)

c) \(...=2.\left(\left(5.16-18\right)+36:18\right)=2.\left(70+2\right)=2.72=144\)

d) \(...=5\left(48:8+44\right)=5\left(6+44\right)=5.50=250\)

e) \(...=20-6=14\) 

f) \(...=50+13-2=61\)

a) 16 . 36 - 16 . 37

= 16 . (36 - 37)

= 16 . (-1)

= -16

b) 36 : 4 . 3 + 9 . 50

= 9 . 3 + 9 . 50

= 9 . (3 + 50)

= 9 . 53

= 477

c) 2 . (5 . 16 - 18) + 36 : 18 . 2

= 2 . (80 - 18) + 2 . 2

= 2 . 62 + 2 . 2

= 2 . (62 + 2)

= 2 . 64

= 128

xét tam giác ABH và Tam giác ACH có :

AC=AB(tính chất tam giác cân)

AHB=AHC(AH vg góc BC)

AH chung

do đó tam giác ABH=tam giác ACH(ch-gn)

b,tAm giác ABC có AH là đường cao xuất phát từ đỉnh A đồng thời là đường phân giác .Suy ra :góc BAH=CAH^(1) HAY EAH^=CAH^

vì EH //AC nên :CAH^=AHE^(2 góc sltrong)(2)

Từ (1) và(2) suy raEAH^=AHE^

suy ra tam giác AHE cân tại E

\(x+y=a\left(1\right)\)

\(x-y=b\left(2\right)\)

\(\left(1\right)+\left(2\right)\Rightarrow2x=a+b\Rightarrow x=\dfrac{a+b}{2}\)

\(\left(1\right)\Rightarrow y=a-x\Rightarrow y=a-\dfrac{a+b}{2}\Rightarrow y=\dfrac{a-b}{2}\)

\(xy=\dfrac{\left(a+b\right)}{2}.\dfrac{\left(a-b\right)}{2}=\dfrac{a^2-b^2}{4}\)

\(x^3-y^3=\left(\dfrac{a+b}{2}\right)^3-\left(\dfrac{a-b}{2}\right)^3=\dfrac{\left(a+b\right)^3}{8}-\dfrac{\left(a-b\right)^3}{8}\)

\(=\dfrac{\left(a+b\right)^3-\left(a-b\right)^3}{8}\)

\(=\dfrac{\left(a+b-a+b\right)\left[\left(a+b\right)^2+\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]}{8}\)

\(=\dfrac{2b\left[a^2+b^2+2ab+a^2-b^2+a^2+b^2-2ab\right]}{8}\)

\(=\dfrac{b\left[3a^2+b^2+2ab\right]}{4}\)

\(\left\{{}\begin{matrix}x+y=a\\x-y=b\end{matrix}\right.\) tính \(x^3\) - y³ theo \(a\) và \(b\)

⇒ \(\left\{{}\begin{matrix}x+y+x-y=a+b\\x-y=b\end{matrix}\right.\)

⇔ \(\left\{{}\begin{matrix}2x=a+b\\y=x-b\end{matrix}\right.\)

⇔ \(\left\{{}\begin{matrix}x=\left(a+b\right):2\\y=\left(a-b\right):2\end{matrix}\right.\) ⇒ \(xy\) = \(\dfrac{a+b}{2}\)\(\times\)\(\dfrac{a-b}{2}\) = \(\dfrac{a^2-b^2}{4}\)

\(x^{3^{ }}\) - y³ = (\(x\) - y)(\(x^2\) + \(x\)y + y²) = \(\left(x-y\right)\)\(\left(\left[x+y\right]^2-xy\right)\) (1)

Thay \(x-y\) = a; \(x\) + y = b và \(xy\) = \(\dfrac{a^2-b^2}{4}\) vào (1) ta có:

\(x^3\) - y³ = b.(a² - \(\dfrac{a^2-b^2}{4}\)) = b.\(\dfrac{3a^2+b^2}{4}\) = \(\dfrac{3a^2b+b^3}{4}\)

(y-240x220)=2020x33

=>y-52800=66660

=>y=66660+52800=119460

` 118/210 - 32/7 = 118/210 - 960/210 = -842/210 = -421/105`

\(\dfrac{118}{210}-\dfrac{32}{7}=-\dfrac{421}{105}\)