1/2 × x +150 ℅ × x = 2004
1/2 × x + 1/8x = 3/4
2- (3/4-x)= 7/12
(2x -4,5):3/4-1/3 =1
Bai 2 tinh gia tri bieu thuc
Q = (1/99+12/999+123/999)×(1/2-1/3-1/6)
℅ la phan tram
Nhanh nhe ai dung minh se thich va keu goi ban be thich nua thank you
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\(\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{999}\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
\(=\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{999}\right).0\)
\(=0\)
\(\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{9999}\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)=\(\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{9999}\right).0=0\)
Bạn ơi đăng từng bài 1 bạn nhé
Lí do:
Bạn đăng thiếu dấu hơi khó nhìn
Hok tốt
a) x + \(\frac{1}{6}\)=\(-\frac{3}{8}\)
x=\(\frac{-3}{8}-\frac{1}{6}\)
x=\(\frac{-13}{24}\)
b)\(\frac{1}{2}\)x + \(\frac{1}{8}\)x = \(\frac{3}{4}\)
\(\left(\frac{1}{2}+\frac{1}{8}\right)\)x = \(\frac{3}{4}\)
\(\frac{5}{8}\)x = \(\frac{3}{4}\)
x = \(\frac{3}{4}:\frac{5}{8}\)
x =\(\frac{6}{5}\)
c) 2 - l3/4-xl=7/12
l3/4-xl=17/12
nếu 3/4 - x > 0
thì 3/4 - x = 17/12
x= -2/3
nếu 3/4 - x <0
thì 3/4 - x = -17/12
x = 13/6
d) \(\left(2x-4.5\right):\frac{3}{4}-\frac{1}{3}=1\)
\(\left(2x-4.5\right):\frac{3}{4}=\frac{4}{3}\)
\(2x-4.5=1\)
2x = 5.5
x = 2.75
a, x + 1/6 = -3/8
x = -3/8 - 1/6
x = -13/24
b, 1/2x + 1/8x = 3/4
x * (1/2+1/8) = 3/4
x * 5/8 = 3/4
x = 3/4 : 5/8
x = 6/5
c, 2 - |3/4-x| = 7/12
|3/4-x| = 2 - 7/12
|3/4-x| = 17/12
=> 3/4 - x = 17/12 hoặc -17/12
# 3/4 - x = 17/12
=> x = -2/3
# 3/4 - x = -17/12
=> x = 13/6
=> x = { -2/3 ; 13/6}
d, (2x - 4,5) : 3/4 -1/3 = 1
(2x - 4,5) : 3/4 = 1 + 1/3
(2x - 4,5 ) : 3/4 = 4/3
2x - 4,5 = 4/3 * 3/4
2x - 4,5 = 1
2x = 1 + 4,5
2x = 5,5
x = 5,5 : 2
x = 2,75
Q=(1/99+12/999+123/999)x(1/6-1/6)
Q=(1/99+12/999+123/999)x0
Q=0
học tốt!
A=(1/99+12/999+123/999)x(1/2-1/3-1/6)
=(1/99+12/999+123/999)x(3/6-2/6-1/6)
=(1/99+12/999+123/999)x0
=0
Bài 1:
a: Để A là phân số thì n+1<>0
hay n<>-1
b: Để A là số nguyên thì \(n+1\in\left\{1;-1;5;-5\right\}\)
hay \(n\in\left\{0;-2;4;-6\right\}\)
x2-4x+7 = 0 ⇔ x2 -4x + 4 + 3 = 0
⇔ (x-2)2+3=0 ⇔ (x-2)2=-3 (vô lí)
Vậy pt vô nghiệm
*Chứng minh phương trình \(x^2-4x+7=0\) vô nghiệm
Ta có: \(x^2-4x+7=0\)
\(\Leftrightarrow x^2-4x+4+3=0\)
\(\Leftrightarrow\left(x-2\right)^2+3=0\)
mà \(\left(x-2\right)^2+3\ge3>0\forall x\)
nên \(x\in\varnothing\)(đpcm)
a,[120-12*5]:6
=[120-60]:6
=60:6
=10
b,494-[75+35:5]*2
=494-[75+7]*2
=494-82*2
=494-164
=330
c,1/4+[3/4:1/2]*5/6
=1/4+3/2*5/6
=1/4+5/4
=3/3
d,[360-15*6]:5
=[360-90]:5
=270:5
=54
a) (120-12*5):6= (120-60):6=60:6=10
b) 494-(75+35:5)*2= 494-(75+7)*2=494-82*2=494-164=330
c)1/4+(3/4:1/2)*5/6=1/4+3/2*5/6=1/4+5/4=6/4=3/2
d)(360-15*6):5=(360-90):5=270:5=54
2,
Q=\(\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{999}\right)x\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
= \(\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{999}\right)x\left(\frac{3+\left(-2\right)+\left(-1\right)}{6}\right)\)
=\(\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{999}\right)x0\)
Vậy Q=0