Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a,\(x+\frac{1}{4}=\frac{3}{4} \)
<=>x=\(\frac{3}{4}-\frac{1}{4} \)
<=>\(x=\frac{1}{2} \)
b,\(\frac{3}{4}-\frac{2}{5}x=\frac{29}{60} \)
<=>\(\frac{2}{5}x =\frac{3}{4}-\frac{29}{60} \)
<=>\(\frac{2}{5}x=\frac{4}{15} \)
<=x=\(\frac{2}{3} \)
c,\(2x-\frac{1}{3}=\frac{-5}{6} \)
2x=\(\frac{-5}{6}+\frac{1}{3} \)
2x=\(\frac{-1}{2} \)
x=\(\frac{-1}{4} \)
d,2-\(\frac{3}{4x}=\frac{1}{2} \)
<=>\(\frac{3}{4x}=2-\frac{1}{2} \)
<=>\(\frac{3}{4}x=\frac{3}{2} \)
<=>x=2
e,\(\frac{11}{12}- \frac{2}{3}|x| =\frac{3}{8} \)
<=>\(\frac{2}{3}|x|=\frac{13}{24} \)
<=>\(|x|=\frac{13}{16} \)
<=>x=\(\pm\frac{13}{16} \)
f,\(|2x-1|=5\)
<=>2x-1=5 hoặc 2x-1=-5
<=> 2x=6 2x=-4
<=> x=3 x=-2
Giải
a) \(x+\dfrac{1}{4}=\dfrac{3}{4}\)
=> \(x=\dfrac{3}{4}-\dfrac{1}{4}\)=>\(x=\dfrac{1}{2}\)
b)\(\dfrac{3}{4}-\dfrac{2}{5}x=\dfrac{29}{60}\)
=>\(\dfrac{2}{5}x=\dfrac{3}{4}-\dfrac{29}{60}\)=>\(\dfrac{2}{5}x=\dfrac{4}{15}\)
=>\(x=\dfrac{4}{15}:\dfrac{2}{5}=\dfrac{2}{3}\)
c)\(2x-\dfrac{1}{3}=\dfrac{-5}{6}\)=>\(2x=\dfrac{-5}{6}+\dfrac{1}{3}\)=\(\dfrac{-1}{2}\)
=>\(x=\dfrac{-1}{2}:2=\dfrac{-1}{4}\)
d)\(2-x:\dfrac{3}{4}=\dfrac{1}{2}\)=>\(x:\dfrac{3}{4}=2-\dfrac{1}{2}\)=\(\dfrac{3}{2}\)
=>\(x=\dfrac{3}{2}.\dfrac{3}{4}=\dfrac{9}{8}\)
e)\(\dfrac{11}{12}-\dfrac{2}{3}.\left|x\right|=\dfrac{3}{8}\)=>\(\dfrac{2}{3}.\left|x\right|=\dfrac{11}{12}-\dfrac{3}{8}\)=\(\dfrac{13}{24}\)
=>\(\left|x\right|=\dfrac{13}{24}:\dfrac{2}{3}=\dfrac{13}{16}\)
Vậy: \(x=\dfrac{13}{16}\)hoặc\(x=\dfrac{-13}{16}\)
f)\(\left|2x-1\right|=5\)
*2x-1=5 =>2x=5+1=6 =>x=6:2=3
*2x-1=-5 =>2x=(-5)+1=-4 =>x=-4:2=-2
Vậy: x=3 hoặc x=-2
Tick cho Phong nhé:>
Yêu nhiều>3
#Phong_419
Noob ơi, bạn phải đưa vào máy tính ý solve cái là ra x luôn, chỉ tội là đợi hơi lâu
a, 4.(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x + 14)
=> 72 - 20x - 36x + 84 = 30x - 240 - 6x - 84
=> (72 + 84) + (-20x - 36x) = (30x - 6x) + (-240 - 84)
=> 156 - 56x = 24x - 324
=> 24x + 56x = 324 + 156
=> 80x = 480
=> x = 480 : 80 = 6
Vậy x = 6
a)Đặt k, ta có:
x/2=k =>2k=x; y/3=k =>3k=y; z/5=k =>5k=z
thay x/2=k =>2k=x; y/3=k =>3k=y; z/5=k =>5k=z vào x2+y2+z2=152, tao có:
(2k)2+(3k)2+(5k)2=152
=>4xk2+9xk2+25xk2=152
=>k2x38=152
=>k2=4=>k=2 hoặc k=-2
Với k=2
=>x=4;y=6;z=10
Với k=-2
=>x=-4;y=-6;z=-10
Vậy (x=4;y=6;z=10) hoặc (x=-4;y=-6;z=-10)
b)Áp dụng dãy tỉ số bằng nhau, ta có :
x/4=y/7=z/9=(2x)/8=(2x-y)/8-7=2
=>x=8;y=14;z=18
Vậy........
a: \(A=\dfrac{5}{4}\cdot\dfrac{11}{3}\cdot\dfrac{-1}{11}=\dfrac{-5}{12}=\dfrac{-25}{60}=\dfrac{-50}{120}\)
b: \(B=\dfrac{3}{4}\cdot\dfrac{1}{12}\cdot\dfrac{2}{3}=\dfrac{1}{24}=\dfrac{5}{120}\)
c: \(C=\dfrac{5}{4}\cdot\dfrac{1}{15}\cdot\dfrac{2}{5}=\dfrac{2}{60}=\dfrac{1}{30}=\dfrac{4}{120}\)
\(D=-3\cdot\dfrac{-7}{12}\cdot\dfrac{1}{-7}=-\dfrac{1}{4}=\dfrac{-30}{120}\)
Vì -50<-30<4<5
nên A<D<B<C
a: \(A=\dfrac{5}{4}\cdot\dfrac{11}{3}\cdot\dfrac{-1}{11}=\dfrac{-5}{12}=\dfrac{-25}{60}=\dfrac{-50}{120}\)
b: \(B=\dfrac{3}{4}\cdot\dfrac{1}{12}\cdot\dfrac{2}{3}=\dfrac{1}{24}=\dfrac{5}{120}\)
c: \(C=\dfrac{5}{4}\cdot\dfrac{1}{15}\cdot\dfrac{2}{5}=\dfrac{2}{60}=\dfrac{1}{30}=\dfrac{4}{120}\)
\(D=-3\cdot\dfrac{-7}{12}\cdot\dfrac{1}{-7}=-\dfrac{1}{4}=\dfrac{-30}{120}\)
Vì -50<-30<4<5
nên A<D<B<C