Tìm x để biểu thức sau có giá trị dương A = x + 27 5 − 3x − 7 4
A. x ≤ 13
B. x > 13
C. x < 13
D. x ≥ 13
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Bài 1: A = (\(\dfrac{7}{13}\) + \(\dfrac{6}{13}\)) x 100 - 13 x a
Thay a = 10 vào A ta có:
A = (\(\dfrac{7}{13}\) + \(\dfrac{6}{13}\)) x 100 - 13 x 10
A = \(\dfrac{13}{13}\) x 100 - 130
A = 100 - 130
A = - 30
Thay a = 987 vào biểu thức A ta có:
A = (\(\dfrac{7}{13}\) + \(\dfrac{6}{13}\)) x 100 - 13 x 987
A = \(\dfrac{13}{13}\) x 100 - 12831
A = 100 - 12831
A = -12731
Ta có:
\(\frac{4}{3\times7}+\frac{5}{7\times12}+\frac{1}{12\times13}+\frac{7}{13\times20}+\frac{3}{20\times23}\)
=>\(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+\frac{1}{13}-\frac{1}{20}+\frac{1}{20}-\frac{1}{23}\)
=>\(\frac{1}{3}-\frac{1}{23}=\frac{20}{69}\)
Chúc bạn học tốt !!!
2,
a) \(315-\left(135-x\right)=215\)
\(\Rightarrow135-x=315-215\)
\(\Rightarrow135-x=100\)
\(\Rightarrow x=135-100\)
\(\Rightarrow x=35\)
b) \(x-320:32=25\cdot16\)
\(\Rightarrow x-10=5^2\cdot4^2\)
\(\Rightarrow x-10=20^2\)
\(\Rightarrow x-10=400\)
\(\Rightarrow x=410\)
c) \(3\cdot x-2018:2=23\)
\(=3\cdot x-1009=23\)
\(\Rightarrow3\cdot x=1032\)
\(\Rightarrow x=1032:3\)
\(\Rightarrow x=344\)
d) \(280-9\cdot x-x=80\)
\(\Rightarrow280-x\cdot\left(9+1\right)=80\)
\(\Rightarrow280-10\cdot x=80\)
\(\Rightarrow10\cdot x=280-80\)
\(\Rightarrow10\cdot x=200\)
\(\Rightarrow x=20\)
e) \(38\cdot x-12\cdot x-x\cdot16=40\)
\(\Rightarrow x\cdot\left(38-12-16\right)=40\)
\(\Rightarrow x\cdot10=40\)
\(\Rightarrow x=40:10\)
\(\Rightarrow x=4\)
T a c ó : A = B ⇔ ( x + 1 ) 3 – ( x – 2 ) 3 = ( 3 x – 1 ) ( 3 x + 1 ) ⇔ x 3 + 3 x 2 + 3 x + 1 – x 3 + 6 x 2 – 12 x + 8 = 9 x 2 – 1 ⇔ x 3 – x 3 + 3 x 2 + 6 x 2 – 9 x 2 + 3 x – 12 x = - 1 – 1 – 8 ⇔ - 9 x = - 10 ⇔ x = 10 / 9 V ậ y v ớ i x = 10 / 9 t h ì A = B .
A=\(\frac{6}{19}\). \(\frac{-7}{11}\)+\(\frac{6}{19}\).\(\frac{-4}{11}\)+\(\frac{-13}{19}\)
=\(\frac{6}{19}\).(\(\frac{-7}{11}\)+\(\frac{-4}{11}\))+\(\frac{-13}{19}\)
=\(\frac{6}{19}\).\(\frac{-11}{11}\)+\(\frac{-13}{19}\)
=\(\frac{6}{19}\).-1 +\(\frac{-13}{19}\)
=\(\frac{-6}{19}\)+\(\frac{-13}{19}\)
=\(\frac{-19}{19}\)
+1