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a. \(\frac{42}{x}+\frac{36}{x}=6\left(x\ne0\right)\Rightarrow\frac{42+36}{x}=6\Leftrightarrow\frac{78}{x}=6\Leftrightarrow x=\frac{78}{6}=13.\)
b. \(\frac{90}{x}-\frac{48}{x}=3\left(x\ne\right)0\Rightarrow\frac{90-48}{x}=3\Leftrightarrow\frac{42}{x}=3\Leftrightarrow x=\frac{42}{3}=14\)
a)42 : x + 36: x=6
(42+36):x=6
78:x =6
x=78:6
x=13
b)90:x-48:x=3
(90-48):x=3
42:x =3
x =42:3
x =14
a) 42 : x + 36 : x = 6
x = ( 42 + 36 ) : x = 6
x = 78 : x = 6
x = 78 : 6
x = 13
b) 90 : x - 48 : x = 3
x = (90 - 48 ) : x = 3
x = 42 : x = 3
x = 42 : 3
x = 14
cho mình một tim nhé
a) 42 : x + 36 : x = 6
( 42 + 36 ) : x = 6
78 : x = 6
x = 78 : 6 = 13
b) 90 : x - 48 : x = 3
( 90 - 48 ) : x = 3
42 : x = 3
x = 42 : 3 = 14
~~~ HT ~~~
a) (42+36):x=66
78:x=66
x=78:66
x=13/11
b) 90:x-48:x=3
(90-48):x=3
42:x=3
x=42:3
x= 14
a) 42:x+36:x=66
42.1/x+36.1/x=66
1/x.(42+36)=66
1/x=11/13
x=1:11/13
x=13/11
b) 90:x-48:x=3
90.1/x-48.1/x=3
1/x(90-48)=3
1/x=1/14
x=14
\(\left(x+2\right)+\left(x+4\right)+\left(x+6\right)+\left(x+8\right)=36\)
\(\Leftrightarrow4x+\left(2+4+6+8\right)=36\)
\(\Leftrightarrow4x+20=36\)
\(\Leftrightarrow x=4\)
Vậy \(x=4\)
\(a,\frac{5}{6}+x\)x \(\frac{3}{4}=3\)
\(x\)x \(\frac{3}{4}=3-\frac{5}{6}\)
\(x\)x \(\frac{3}{4}=\frac{13}{6}\)
\(x=\frac{13}{6}:\frac{3}{4}\)
\(x=\frac{26}{9}\)
b, \(5\frac{1}{6}-x:\frac{3}{4}=\frac{2}{3}\)
\(\frac{31}{6}-x:\frac{3}{4}=\frac{2}{3}\)
\(x:\frac{3}{4}=\frac{31}{6}-\frac{2}{3}\)
\(x:\frac{3}{4}=\frac{9}{2}\)
\(x=\frac{9}{2}\)x \(\frac{3}{4}\)
\(x=\frac{27}{8}\)
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!Thank you!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
\(\left(2.8x-32\right):\frac{2}{3}=90\)
\(2.8\cdot x-32=90\cdot\frac{2}{3}\)
\(\frac{14}{5}x-32=60\)
\(\frac{14}{5}x=60+32\)
\(\frac{14}{5}x=92\)
\(x=\frac{230}{7}\)
B , c , d tương tự
\(x+6x=42\)
\(\Leftrightarrow7x=42\)
\(\Leftrightarrow x=42:7\)
\(\Leftrightarrow x=6\)
Vậy ...
\(x^2=x\)
\(\Leftrightarrow x^2-x=0\)
\(\Leftrightarrow x\left(x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x-1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
Vậy ...
\(x+6x=42\)
\(\Leftrightarrow7x=42\)
\(\Rightarrow x=6\)
hok tốt
Bài 1 :
\(a)\) Ta có :
\(3x=4y=6z\)
\(\Leftrightarrow\)\(\frac{3x}{12}=\frac{4y}{12}=\frac{6z}{12}\)
\(\Leftrightarrow\)\(\frac{x}{4}=\frac{y}{3}=\frac{z}{2}\)
\(\Leftrightarrow\)\(\frac{2x}{8}=\frac{y}{3}=\frac{5z}{10}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{2x}{8}=\frac{y}{3}=\frac{5z}{10}=\frac{2x-5z}{8-10}=\frac{-36}{-2}=18\)
Do đó :
\(\frac{x}{4}=18\)\(\Rightarrow\)\(x=18.4=72\)
\(\frac{y}{3}=18\)\(\Rightarrow\)\(y=18.3=54\)
\(\frac{z}{2}=18\)\(\Rightarrow\)\(z=18.2=36\)
Vậy \(x=72\)\(;\)\(y=54\) và \(z=36\)
Chúc bạn học tốt ~
2) Ta có: \(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=\frac{a+b+c}{b+c+c+a+a+b}=\frac{a+b+c}{2.\left(a+b+c\right)}=\frac{1}{2}\)
\(\Rightarrow\frac{a}{b+c}=\frac{1}{2}\Rightarrow2a=b+c\)
\(\frac{b}{c+a}=\frac{1}{2}\Rightarrow2b=c+a\)
\(\frac{c}{a+b}=\frac{1}{2}\Rightarrow2c=a+b\)
Ta có: \(\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\frac{b+a}{b}.\frac{c+b}{c}.\frac{a+c}{a}=\frac{2c.2a.2b}{b.c.a}=8\)
Vậy \(\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=8\)