Bài 1: Tính biểu thức, biết:
( 10 + 20 + 30 +... + 100 ) x ( a x 1 - a : 1 )
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Bài 1 :
\(M=\dfrac{30-2^{20}}{2^{18}}=\dfrac{2.15-2^{20}}{2^{18}}=\dfrac{15}{2^{17}}-2^2=\dfrac{15}{2^{17}}-4< 0\left(\dfrac{15}{2^{17}}< 1\right)\)
\(N=\dfrac{3^5}{1^{2021}+2^3}=\dfrac{3^5}{9}=\dfrac{3^5}{3^2}=3^3=27\)
\(\Rightarrow M< N\)
Bài 3 :
a) \(t^2+5t-8\) khi \(t=2\)
\(=5^2+2.5-8\)
\(=25+10-8\)
\(=27\)
b) \(\left(a+b\right)^2-\left(b-a\right)^3+2021\left(1\right)\)
\(\left\{{}\begin{matrix}a=5\\b=a+1=6\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b=11\\b-a=1\end{matrix}\right.\)
\(\left(1\right)=11^2-1^3+2021=121-1+2021=2141\)
c) \(x^3-3x^2y+3xy^2-y^3=\left(x-y\right)^3\left(1\right)\)
\(\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\) \(\Rightarrow x-y=1\)
\(\left(1\right)=1^3=1\)
a) \(\frac{x^7}{3^5}=9\)
=> \(\frac{x^7}{3^5}=3^2\)
=> x7 = 32 . 35 = 37
=> x = 3
b) \(\frac{32}{x^7}=4\)
=> \(\frac{2^5}{x^7}=2^2\)
=> \(2^2\cdot x^7=2^5\)
=> \(x^7=\frac{2^5}{2^2}=2^3\)
=> không tìm được x
2. \(\frac{2^{30}}{3^{20}}-\left(\frac{2}{3}\right)^{20}\cdot2^{10}\)
\(=\frac{2^{30}}{3^{20}}-\frac{2^{20}}{3^{20}}\cdot2^{10}\)
\(=\frac{2^{30}}{3^{20}}-\frac{2^{30}}{3^{20}}=0\)
\(a,A=\left(x+10\right)+\left(2x-15\right)-\left(x-20\right)\)
\(=x+10+2x-15-x+20\)
\(=2x+15\)
\(b,\)Thay \(x=15\)vào biểu thức A = 2x + 15
Ta được : \(A=2\times15+15\)
\(\Rightarrow A=45\)
a) A = (x + 10) + (2x - 15) - (x - 20)
A = x + 10 + 2x - 15 - x + 20
A = x + 10 + 2x + (-15) + (-x) + 20
A = x + (-x) + 2x + 10 + 20 + (-15)
A = 2x + 15
b) x = 15 thì A = 2x + 15 = 2.15 + 15 = 15.3 = 45
Đặt \(g(x)=10x\).
Ta có \(g\left(1\right)=10=f\left(1\right);g\left(2\right)=20=f\left(2\right);g\left(3\right)=30=f\left(3\right)\).
Từ đó \(\left\{{}\begin{matrix}f\left(1\right)-g\left(1\right)=0\\f\left(2\right)-g\left(2\right)=0\\f\left(3\right)-g\left(3\right)=0\end{matrix}\right.\)
\(\Rightarrow f\left(x\right)-g\left(x\right)=Q\left(x\right).\left(x-1\right)\left(x-2\right)\left(x-3\right)\).
\(\Rightarrow f\left(x\right)=10x+Q\left(x\right)\left(x-1\right)\left(x-2\right)\left(x-3\right)\)
\(\Rightarrow f\left(8\right)+f\left(-4\right)=80+Q\left(x\right).7.6.5+\left(-40\right)+Q\left(x\right).\left(-5\right).\left(-6\right).\left(-7\right)=80-50=40\).
Đoạn cuối mình làm nhầm nhé.
Đáng lẽ phải cm Q(x) là đa thức dạng x + m, rồi biến đổi \(f\left(8\right)+f\left(-4\right)=80+Q\left(8\right).7.6.5+\left(-40\right)+Q\left(-4\right).\left(-5\right).\left(-6\right).\left(-7\right)=80-40+\left(m+8\right).7.6.5-\left(m-4\right).5.6.7=12.5.6.7+40=2560\).
Mình đánh vội nên chưa suy nghĩ kĩ.
1)
a) (a-b) - (c-b) - a= a - b - c + b - a = -c
b) -(300 - 400) + (300 - 400) + 100 = 100
2) a) 4x - 20 = 50 - ( 30 -3x)
4x - 20 = 50 - 30 + 3x
4x - 3x = 50 - 30 + 20= 40
x = 40
b) 100 - (-5x) = 40 - (-4x +10)
100 + 5x = 40 + 4x -10
5x - 4x = 40 - 10 -100
x = -70
c) (-50) + (-7x) = 100 - 8x
-7x + 8x = 100 + 50
x = 150
Đặt \(g\left(x\right)=f\left(x\right)-10\) (bậc 4)
\(\Leftrightarrow\left\{{}\begin{matrix}g\left(1\right)=0\\g\left(2\right)=0\\g\left(3\right)=0\end{matrix}\right.\Leftrightarrow g\left(x\right)=\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-m\right)\) (m là hằng số)
\(\Leftrightarrow f\left(x\right)=\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-m\right)-10\\ \Leftrightarrow f\left(9\right)=8\cdot7\cdot6\left(9-m\right)-10=336\left(9-m\right)-10\\ f\left(-5\right)=\left(-6\right)\left(-7\right)\left(-8\right)\left(-5-m\right)-10=336\left(m+5\right)-10\)
Vậy \(A=336\left(9-m\right)+336\left(m+5\right)-20=4684\)
Chúc bạn hok tốt <3
Bài 1:
6a+1 \(⋮\)2a-1
=> 2a-1\(⋮\)2a-1
=> (6a+1)- 3(2a-1) \(⋮\)2a-1
=> (6a+1) - ( 6a-3) \(⋮\)2a-1
=> 6a+1 -6a+3\(⋮\)2a-1
=> 4 \(⋮\)2a-1
=> 2a-1\(\in\)Ư(4)
Còn j bn làm nốt nhoaaa
1. Ta có \(6a+1⋮2a-1\)
\(\Rightarrow3\left(2a-1\right)+4⋮a-1\)
\(\Rightarrow4⋮a-1\)
\(\Rightarrow a-1\inƯ\left(4\right)=\left\{-4;-2;-1;1;2;4\right\}\)
\(\Rightarrow a\in\left\{-3;-1;0;2;3;5\right\}\) ( thỏa mãn a nguyên )
Vậy \(a\in\left\{-3;-1;0;2;3;5\right\}\)
2. a, x - y - ( - y + a + x)
= x - y + y - a - x
= - a
b, (-90) - (b + 10) + 100
= - 90 - b - 10 + 100
= ( - 90 - 10 +100) - b
= 0 - b
= - b
@@ Học tốt
( 10 + 20 + 30 +... + 100 ) x ( a x 1 - a : 1 )
( 10 + 20 + 30 +... + 100 ) x ( a - a)
( 10 + 20 + 30 +... + 100 ) x 0
= 0