C=1×5+2×6+3×7+...+20×24
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a) \(\dfrac{13}{20}+\dfrac{3}{5}+x=\dfrac{5}{6}\)
\(\Rightarrow\dfrac{5}{4}+x=\dfrac{5}{6}\)
\(\Rightarrow x=\dfrac{5}{6}-\dfrac{5}{4}\)
\(\Rightarrow x=\dfrac{-5}{12}\)
b) \(x+\dfrac{1}{3}=\dfrac{2}{5}-\dfrac{-1}{3}\)
\(\Rightarrow x+\dfrac{1}{3}=\dfrac{11}{15}\)
\(\Rightarrow x=\dfrac{11}{15}-\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{2}{5}\)
c)\(\dfrac{-5}{8}-x=\dfrac{-3}{20}-\dfrac{-1}{6}\)
\(\dfrac{-5}{8}-x=\dfrac{1}{60}\)
\(\Rightarrow x=\dfrac{-5}{8}-\dfrac{1}{60}\)
\(\Rightarrow x=\dfrac{-77}{120}\)
d) \(\dfrac{3}{5}-x=\dfrac{1}{4}+\dfrac{7}{10}\)
\(\Rightarrow\dfrac{3}{5}-x=\dfrac{19}{20}\)
\(\Rightarrow x=\dfrac{3}{5}-\dfrac{19}{20}\)
\(\Rightarrow x=\dfrac{-7}{20}\)
e) \(\dfrac{-3}{7}-x=\dfrac{4}{5}+\dfrac{-2}{3}\)
\(\Rightarrow\dfrac{-3}{7}-x=\dfrac{2}{15}\)
\(\Rightarrow x=\dfrac{-3}{7}-\dfrac{2}{15}\)
\(\Rightarrow x=\dfrac{-59}{105}\)
g) \(\dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)
\(\Rightarrow\dfrac{-5}{6}-x=\dfrac{1}{4}\)
\(\Rightarrow x=\dfrac{-5}{6}-\dfrac{1}{4}\)
\(\Rightarrow x=\dfrac{-13}{12}\)
Giải:
a) \(2\dfrac{17}{20}-1\dfrac{15}{11}+6\dfrac{9}{20}:3\)
\(=\dfrac{57}{20}-\dfrac{26}{11}+\dfrac{129}{20}:3\)
\(=\dfrac{107}{220}+\dfrac{43}{20}\)
\(=\dfrac{29}{11}\)
b) \(4\dfrac{3}{7}:\left(\dfrac{7}{5}.4\dfrac{3}{7}\right)\)
\(=\dfrac{31}{7}:\left(\dfrac{7}{5}.\dfrac{31}{7}\right)\)
\(=\dfrac{31}{7}:\dfrac{31}{5}\)
\(=\dfrac{5}{7}\)
c) \(\left(3\dfrac{2}{9}.\dfrac{15}{23}.1\dfrac{7}{29}\right):\dfrac{5}{23}\)
\(=\left(\dfrac{29}{9}.\dfrac{15}{23}.\dfrac{36}{29}\right):\dfrac{5}{23}\)
\(=\dfrac{60}{23}:\dfrac{5}{23}\)
\(=12\)
\(\dfrac{5}{3}+\dfrac{3}{x}=\dfrac{49}{24}\\ \dfrac{3}{x}=\dfrac{49}{24}-\dfrac{5}{3}\\ \dfrac{3}{x}=\dfrac{3}{8}\\ x=8\)
\(\dfrac{3}{x}=\dfrac{49}{24}-\dfrac{5}{3}\\ \dfrac{3}{x}=\dfrac{49}{24}-\dfrac{40}{24}\\ \dfrac{3}{x}=\dfrac{3}{8}\\ x=8\)
a) \(\left(x+1\right)\left(x+2\right)\left(x+4\right)\left(x+5\right)-4=\left(x^2+6x+5\right)\left(x^2+6x+8\right)-4\)
Đặt \(t=x^2+6x+5\)
\(PT=t\left(t+3\right)-4=t^2+3t-4=\left(t-1\right)\left(t+4\right)\)
Thay t: \(PT=\left(x^2+6x+5-1\right)\left(x^2+6x+5+4\right)=\left(x^2+6x+4\right)\left(x^2+6x+9\right)=\left(x^2+6x+4\right)\left(x+3\right)^2\)
b) Đặt \(t=\left(2x+1\right)^2\)
\(PT=t^2-3t+2=\left(t^2-3t+\dfrac{9}{4}\right)-\dfrac{1}{4}=\left(t+\dfrac{3}{2}\right)^2-\dfrac{1}{4}=\left(t+1\right)\left(t+2\right)\)
Thay t:
\(PT=\left[\left(2x+1\right)^2+1\right]\left[\left(2x+1\right)^2+2\right]=\left[4x^2+4x+2\right]\left[4x^2+4x+3\right]=2\left[2x^2+2x+1\right]\left[4x^2+4x+3\right]\)
a) (38 - 60) + (20 - 38)
= 38 - 60 + 20 - 38
= (38 - 38) + (-60 + 20)
= 0 - 40
= -40
b) 75 - (20 + 75)
= 75 - 20 - 75
= (75 - 75) - 20
= 0 - 20
= -20
c) 32 + (60 - 32)
= 32 + 60 - 32
= (32 - 32) + 60
= 0 + 60
= 60
d) (81 - 36) - (81 - 36)
= 81 - 36 - 81 + 36
= (81 - 81) + (-36 + 36)
= 0 + 0
= 0
e) (2 + 4 + 6 + 8) - (1 + 3 + 5 + 7)
= 2 + 4 + 6 + 8 - 1 - 3 - 5 - 7
= (2 - 1) + (4 - 3) + (6 - 5) + (8 - 7)
= 1 + 1 + 1 + 1
= 4
f) (1 + 3 + 5 + 7 + ... + 99) - (2 + 4 + 6 + 8 + ... + 100)
= 1 + 3 + 5 + 7 + ... + 99 - 2 - 4 - 6 - 8 - ... - 100
= (1 - 2) + (3 - 4) + (5 - 6) + (7 - 8) + ... + (99 - 100)
= -1 - 1 - 1 - 1 - ... - 1 (50 chữ số 1)
= -50
B1:
A.1+1=2
B.6+2+5=13
C.6+4+6+2+7=25
D.6+4+7+8=25
B2:
A.x+2=3
x=3-2
x=1
B.x+5=4+6
x+5=10
x=10-5
x=5
C.7=4+x
x=7-4
x=3
D.4+x={3+5+7}+{3+5}
4+x=15+8
4+x=23
x=23-4
x=19
Bài 1 :
A.1+1=2
B.6+2+5=13
C.6+4+6+2+7=25
D.6+4+7+8=25
Bài 2 :
A.x+2=3
x=3-2
x=1
B.x+5=4+6
x+5=10
x=10-5
x=5
C.7=4+x
x=7-4
x=3
D.4+x={3+5+7}+{3+5}
4+x=15+8
4+x=23
x=23-4
x=19
Chúc bạn học tốt nha ~_~