\(\dfrac{1}{3}+x=\dfrac{7}{9}\left(x+\dfrac{3}{4}\right)\)
\(\dfrac{1}{3}+x=\dfrac{7}{9}x+\dfrac{7}{12}\)
\(\dfrac{2}{9}x=\dfrac{1}{4}\)
x=\(\dfrac{9}{8}\)
Tìm x, biết :
\(a.3\dfrac{4}{5}:40\dfrac{8}{15}=0,25:x\)
\(b.\left(x+1\right):\dfrac{5}{6}=20:3\)
\(c.\dfrac{7}{x-1}=\dfrac{x+1}{9}\)
tỉ lệ thức
( x- 6 ) : ( -1,5) = -6 : ( x- 6 )
\(1\dfrac{2}{3}:\left(x-4\right)=5\dfrac{1}{6}:\left(x+1\right)\)
tìm x,y thuộc Z biết:
a, \(\left(x+4\right)\left(y+3\right)=3\)
b,\(\left(x+2\right)\left(y-3\right)=-3\)
c,\(\dfrac{x+1}{2}=\dfrac{1}{y}\)
d, \(\dfrac{x-7}{-1}=\dfrac{13}{2-y}\)
tìm x ∈ Q biết : \(\dfrac{3\left(x+1\right)}{2}\)=\(\dfrac{8}{27\left(x-1\right)}\)
Tìm x trong tỉ lệ thức
a) \(-0,52:x=-9,36:16,38\)
b) \(2\dfrac{2}{5}:x=1\dfrac{7}{9}:0,2\)
c)\(\left(\dfrac{2}{5}x\right):\dfrac{1}{3}=1\dfrac{1}{3}:\dfrac{4}{5}\)
d) \(0,3:1\dfrac{1}{3}=\dfrac{2}{5}:\left(3x+6\right)\)
e) \(\dfrac{x}{-15}=\dfrac{-60}{x}\)
f) \(\dfrac{x-1}{-15}=\dfrac{-60}{x-1}\)
g) \(\dfrac{x-1}{7}=\dfrac{9}{x+1}\)
h) \(\dfrac{x-2}{x-1}=\dfrac{x+4}{x+7}\)
1.Tìm x trong các tỉ lệ thức sau:
a,\(16:x^2+x:\left(-4\right)\)
b,\(x:0,27=3:x\)
c,\(0,81:\dfrac{x}{2}=\dfrac{16}{x^4}:\left(-0,9\right)\)
d,\(\dfrac{x-1}{x+2}=\dfrac{x-2}{x+3}\)
e,\(\left(x+15\right):x=\dfrac{4}{3}\)
f,\(\left(x-20\right):\left(x-10\right)=\left(x+40\right):\left(x+70\right)\)
2.Tìm số nguyên x mà khi cùng thêm vào tử và mẫu của phân số \(\dfrac{26}{39}\) ta được phân số mới có giá trị bằng \(\dfrac{6}{7}\)
3. Cho \(\dfrac{a+5}{a-5}=\dfrac{b+6}{b-6}\) (với a≠5, b≠6). Chứng minh rằng \(\dfrac{a}{b}=\dfrac{5}{6}\)
1) \(\dfrac{-5}{\dfrac{12}{\left|\dfrac{2}{3}x+\dfrac{1}{2}\right|}}\)= \(\dfrac{\dfrac{-4}{9}}{\dfrac{8}{15}}\)
2) \(\dfrac{\left|3x-5\right|}{-3}\)= \(\dfrac{\dfrac{-5}{6}}{\dfrac{4}{9}}\)
Tìm \(x\) trong các tỉ lệ thức sau :
a) \(3,8:\left(2x\right)=\dfrac{1}{4}:2\dfrac{2}{3}\)
b) \(\left(0,25x\right):3=\dfrac{5}{6}:0,125\)
c) \(0,01:2,5=\left(0,75x\right):0,75\)
d) \(1\dfrac{1}{3}:0,8=\dfrac{2}{3}:\left(0,1x\right)\)
\(\left|2x+3\right|+\left|2x-1\right|=\dfrac{8}{3.\left(x+1\right)^2+2}\)\(\sqrt{ }\)\(\left|2x+3\right|+\left|2x-1\right|\)=\(\dfrac{8}{3.\left(x+1\right)^2+2}\)
