a) (2x-1)6 = (2x-1)8
b) (4x-3)4 = (4x-3)10
c)23x + 2 = 9x+5
d)\(\frac{1}{9}:27^n=3^n\)
e)\(\frac{2^n}{2}+4.2n=2^{88}\)
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24 tháng 1 2022
d: =>4x+6=15x-12
=>4x-15x=-12-6=-18
=>-11x=-18
hay x=18/11
e: =>\(45x+27=12+24x\)
=>21x=-15
hay x=-5/7
f: =>35x-5=96-6x
=>41x=101
hay x=101/41
g: =>3(x-3)=90-5(1-2x)
=>3x-9=90-5+10x
=>3x-9=10x+85
=>-7x=94
hay x=-94/7
12 tháng 8 2021
d) \(4x^4-x^2=x^2\left(4x^2-1\right)=x^2\left(2x-1\right)\left(2x+1\right)\)
e) Ta có: \(6x^2-7x-5\)
\(=6x^2-10x+3x-5\)
\(=2x\left(3x-5\right)+\left(3x-5\right)\)
\(=\left(3x-5\right)\left(2x+1\right)\)
f: Ta có: \(-4x^2+23x-15\)
\(=-4x^2+20x+3x-15\)
\(=-4x\left(x-5\right)+3\left(x-5\right)\)
\(=\left(x-5\right)\left(-4x+3\right)\)
A
22 tháng 10 2017
4.a)n2(n+1)+2n(n+1)=(n+1)(n2+2n)=n(n+1)(n+2)
n,(n+1),(n+2) là ba số nguyên liên tiếp nên chia hết cho 2 và 3
\(\Rightarrow\)n(n+1)(n+2) chia hết cho 6
22 tháng 10 2017
4 Chứng minh rằng:
a)\(n^2+\left(n+1\right)+2n\left(n+1\right)\) chia hết cho 6
Ta có:
\(n^2\left(n+1\right)+2n\left(n+1\right)\)
\(=n^3+3n^2+2n\)
\(=n\left(n^2+3n+2\right)\)
\(=n\left(n+1\right)\left(n+2\right)\)
Ta thấy n , n+1 và n+2 là ba số tự nhiên liên tiếp
=> n(n+1) (n+2)\(⋮\)6
=> đpcm
b)\(\left(2n-1\right)^3-\left(2n-1\right)\) chia hết cho 8
Ta có:
\(\left(2n-1\right)^3-\left(2n-1\right)\)
\(=\left(2n-1\right)\left[\left(2n-1\right)^2-1\right]\)
\(=\left(2n-1\right)\left[\left(2n-1\right)^2-1^2\right]\)
\(=\left(2n-1\right)\left(2n-1-1\right)\left(2n-1+1\right)\)
\(=\left(2n-1\right).2\left(n-1\right).2n\)
\(=4n\left(2n-1\right)\left(n-1\right)\)
=>\(4n\left(2n-1\right)\left(n-1\right)⋮4\left(1\right)\)
Mà(2n-1)(n-1)=(n+n-1)(n-1)
=>\(\left(2n-1\right)\left(n-1\right)⋮2\left(2\right)\)
Từ (1) và (2)=> Đpcm
c)\(\left(n+7\right)^2-\left(n-5\right)^2\) chia hết cho 24
Câu hỏi của Ngoc An Pham - Toán lớp 8 | Học trực tuyến
Chúc bạn học tốt!^^
9 tháng 6 2020
a, 2(4x - 7 ) = 3(x + 1) + 18
⇌ 8x -14 = 3x + 3 + 18
⇌ 5x = 35 ⇌ x = 7
→ S = \(\left\{7\right\}\)
b, ( 2x - 1 )2 - 4x ( x - 3 ) = -11
⇌ 4x2 - 2x + 1 - 4x2 + 12 = -11
⇌ 10x = -12
⇌ x = \(-\frac{12}{10}\)
→ S = \(\left\{-\frac{12}{10}\right\}\)
c, ( 2x - 5 )2 - ( x + 2 )2 = 0
⇌ ( 2x - 5 -x + 2 )2 = 0
⇌ ( x - 3 )2 = 0
⇌ x - 3 = 0 ⇌ x = 3
→ S = \(\left\{3\right\}\)
d, ( x - 6 ) ( x + 1 ) = 2(x + 1)
⇌ ( x - 6 - 2 ) ( x+ 1) = 0
⇌ x2 - 7x - 8 =0
⇌ ( x - 8 ) ( x + 1 ) = 0
⇒\(\left\{{}\begin{matrix}x-8=0\\x+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=8\\x=-1\end{matrix}\right.\)
→ S = \(\left\{8;-1\right\}\)
e, \(\frac{x-3}{2}=2-\frac{1-2x}{5}\)
⇌ 5( x - 3) = 20 - 2(1 - 2x)
⇌ 5x - 4x = 15 + 20 + 2
⇌ x = 37
→ S = \(\left\{37\right\}\)
g, \(\frac{3x+2}{2}+\frac{5-2x}{3}=\frac{11}{6}\)
⇌ 3(3x + 2) + 2(5 - 2x) = 11
⇌ 6x + 6 + 10 - 4x = 11
⇌ 2x = -5
⇌ x = \(-\frac{5}{2}\)
→ S = \(\left\{-\frac{5}{2}\right\}\)
h, \(\frac{x-2}{x+2}-\frac{3}{x-2}=\frac{9x-66}{x^2-4}\)
⇌ (x - 2)2 - 3(x - 2) = 9x - 66
⇌ x2 - 4x + 4 - 3x - 6 = 9x - 66
⇌ x2 -16 + 64 = 0
⇌ (x - 8)2 = 0
⇌ x - 8 = 0
⇌ x = 8
→ S = \(\left\{8\right\}\)
16 tháng 12 2020
Bài 1.
a)\(\frac{4x-4}{x^2-4x+4}\div\frac{x^2-1}{\left(2-x\right)^2}=\frac{4\left(x-1\right)}{\left(x-2\right)^2}\div\frac{\left(x-1\right)\left(x+1\right)}{\left(x-2\right)^2}=\frac{4\left(x-1\right)}{\left(x-2\right)^2}\times\frac{\left(x-2\right)^2}{\left(x-1\right)\left(x+1\right)}=\frac{4}{x+1}\)
b) \(\frac{2x+1}{2x^2-x}+\frac{32x^2}{1-4x^2}+\frac{1-2x}{2x^2+x}=\frac{2x+1}{x\left(2x-1\right)}+\frac{-32x^2}{4x^2-1}+\frac{1-2x}{x\left(2x+1\right)}\)
\(=\frac{\left(2x+1\right)\left(2x+1\right)}{x\left(2x-1\right)\left(2x+1\right)}+\frac{-32x^3}{x\left(2x-1\right)\left(2x+1\right)}+\frac{\left(1-2x\right)\left(2x-1\right)}{x\left(2x-1\right)\left(2x+1\right)}\)
\(=\frac{4x^2+4x+1}{x\left(2x-1\right)\left(2x+1\right)}+\frac{-32x^3}{x\left(2x-1\right)\left(2x+1\right)}+\frac{-4x^2+4x-1}{x\left(2x-1\right)\left(2x+1\right)}\)
\(=\frac{4x^2+4x+1-32x^3-4x^2+4x-1}{x\left(2x-1\right)\left(2x+1\right)}=\frac{-32x^3+8x}{x\left(2x-1\right)\left(2x+1\right)}\)
\(=\frac{-8x\left(4x^2-1\right)}{x\left(2x-1\right)\left(2x+1\right)}=\frac{-8x\left(2x-1\right)\left(2x+1\right)}{x\left(2x-1\right)\left(2x+1\right)}=-8\)
c) \(\left(\frac{1}{x+1}+\frac{1}{x-1}-\frac{2x}{1-x^2}\right)\times\frac{x-1}{4x}\)
\(=\left(\frac{1}{x+1}+\frac{1}{x-1}+\frac{2x}{x^2-1}\right)\times\frac{x-1}{4x}\)
\(=\left(\frac{x-1}{\left(x-1\right)\left(x+1\right)}+\frac{x+1}{\left(x-1\right)\left(x+1\right)}+\frac{2x}{\left(x-1\right)\left(x+1\right)}\right)\times\frac{x-1}{4x}\)
\(=\left(\frac{x-1+x+1+2x}{\left(x-1\right)\left(x+1\right)}\right)\times\frac{x-1}{4x}\)
\(=\frac{4x}{\left(x-1\right)\left(x+1\right)}\times\frac{x-1}{4x}=\frac{1}{x+1}\)
Bài 3.
N = ( 4x + 3 )2 - 2x( x + 6 ) - 5( x - 2 )( x + 2 )
= 16x2 + 24x + 9 - 2x2 - 12x - 5( x2 - 4 )
= 14x2 + 12x + 9 - 5x2 + 20
= 9x2 + 12x + 29
= 9( x2 + 4/3x + 4/9 ) + 25
= 9( x + 2/3 )2 + 25 ≥ 25 > 0 ∀ x
=> đpcm
nhé
a)(2x-1)6=(2x-1)8
=> (2x-1)8-(2x-1)6=0
=> (2x-1)6.((2x-1)2-1)=0
+)th1(2x-1)6=0
+)th2((2x-1)2-1)=0
a) \(\left(2x-1\right)^6=\left(2x-1\right)^8\)
\(\Rightarrow\left(2x-1\right)\in\left\{\pm1;0\right\}\)
TH1 : \(2x-1=0\) TH2 : \(2x-1=-1\) TH3 : \(2x-1=1\)
\(2x=1\) \(2x=0\) \(2x=2\)
\(x=\frac{1}{2}\) \(x=0\) \(x=1\)
Vậy \(x\in\left\{\frac{1}{2};0;1\right\}\)
b) Tương tự