Tìm x :
16x + 40 = 10 . 3 mũ 2 + 5 . ( 1 + 2 + 3 )
16x+40=10 . 3 mũ 2 + 5 . (1+2+3)
\(16x+40=10\cdot3^2+5\cdot\left(1+2+3\right)\)
\(\Rightarrow16x+40=10\cdot9+5\cdot6\)
\(\Rightarrow16x+40=90+30\)
\(\Rightarrow16x+40=120\)
\(\Rightarrow16x=120-40\)
\(\Rightarrow16x=80\)
\(\Rightarrow x=\dfrac{80}{16}\)
\(\Rightarrow x=5\)
phân tích đa thức thành nhân tử
6, x mũ 2 - 1 + 2xy + y mũ 2
7, 4x mũ 2 - 12x + 9 - y mũ 2
8, 16x mũ 2 - 4y mũ 2 + 4y - 1
9, 25 - x mũ 2 - 12x - 36
10, x mũ 2 - 9 - 5 ( x+ 3 )
6, \(x^2-1+2xy+y^2=\left(x+y\right)^2-1=\left(x+y-1\right)\left(x+y+1\right)\)
7, \(4x^2-12x+9-y^2=\left(2x-3\right)^2-y^2=\left(2x-3-y\right)\left(2x-3+y\right)\)
8, \(16x^2-4y^2+4y-1=16x^2-\left(2y-1\right)^2=\left(4x-2y+1\right)\left(4x+2y-1\right)\)
9, \(25-x^2-12x-36=25-\left(x+6\right)^2=\left(5-x-6\right)\left(5+x+5\right)=-\left(x+1\right)\left(x+10\right)\)
10, \(x^2-9-5\left(x+3\right)=\left(x-3\right)\left(x+3\right)-5\left(x+3\right)=\left(x+3\right)\left(x-8\right)\)
phân tích đa thức sau thành nhân tử
7, 4x mũ 2 - 12x + 9 - y mũ 2
8, 16x mũ 2 - 4y mũ 2 + 4y - 1
9, 25 - x mũ 2 - 12x - 36
10, x mũ 2 - 9 - 5 ( x + 3 )
7, 4x mũ 2 - 12x + 9 - y mũ 2 = -(y-2x+3) (y+2x-3)
8, 16x mũ 2 - 4y mũ 2 + 4y - 1 = -(2y - 4x - 1) (2y+4x-1)
9, 25 - x mũ 2 - 12x - 36 = -(x+1) (x+11)
10, x mũ 2 - 9 - 5 ( x + 3 ) = (x-8) (x+3)
bạn k cho mình nha
chúc bạn học tốt :))))
bạn kham khảo link, mình đã làm rồi nhé
Câu hỏi của Phạm Đỗ Bảo Ngọc - Toán lớp 8 - Học trực tuyến OLM
Trả lời:
7, 4x2 - 12x + 9 - y2
= ( 4x2 - 12x + 9 ) - y2
= ( 2x - 3 )2 - y2
= ( 2x - 3 - y )( 2x - 3 + y )
8, 16x2 - 4y2 + 4y - 1
= 16x2 - ( 4y2 - 4y + 1 )
= ( 4x )2 - ( 2y - 1 )2
= ( 4x - 2y + 1 )( 4x + 2y - 1 )
9, 25 - x2 - 12x - 36
= 25 - ( x2 + 12x + 36 )
= 52 - ( x + 6 )2
= ( 5 - x - 6 )( 5 + x + 6 )
= ( - 1 - x )( 11 + x )
10, x2 - 9 - 5 ( x + 3 )
= ( x - 3 )( x + 3 ) - 5 ( x + 3 )
= ( x + 3 )( x - 3 - 5 )
= ( x + 3 )( x - 8 )
1) 16 mũ 11 x 5 mũ 40/ 10 mũ 41
2) 3 mũ 7 x 8 mũ 5/ 6 mũ 6 x(-2) mũ 12
1: \(\dfrac{16^{11}\cdot5^{40}}{10^{41}}=\dfrac{2^{44}\cdot5^{40}}{2^{41}\cdot5^{41}}=\dfrac{2^3}{5^1}=\dfrac{8}{5}\)
2: \(\dfrac{3^7\cdot8^5}{6^6\cdot\left(-2\right)^{12}}=\dfrac{3^7\cdot2^{15}}{2^6\cdot3^6\cdot2^{12}}=\dfrac{3}{2^3}=\dfrac{3}{8}\)
12+(5+x)=20
5.2 mũ 2+(x+3)=5 mũ 2
2 mũ 3 +(x+3)=5 mũ 2
4 mũ 3 -(x-2)=5 mũ 2
7 nhân ( x +5 ) + 14 = 7 mũ 3
14 x - 2 nhân 7 mũ 2=2 nhân 3 nhân 7
6 nhân ( x + 2 mũ 3 ) + 40 = 100
155 - 10 nhan ( x +1)=55
12 + ( 5 + x ) = 20 5.22 + ( x + 3 ) = 52 23 + ( x + 3 ) = 52 43 - ( x - 2 ) = 52
17 + x = 20 5.4 + x + 3 = 25 8 + x + 3 = 25 64 - x + 2 = 25
x = 20 - 17 20 + 3 + x = 25 11 + x = 25 66 - x = 25
x = 3 23 + x = 25 x = 25 - 11 x = 66 - 25
x = 25 - 23 x = 14 x = 41
x = 2
Đăng nhìu v bn :) Đáng quan ngại đây :)
1. 6 X mũ 3 -8 =40
2. 4 X mũ 5 +15=47
3. 2 X mũ 3-4=12
4. 5 X mũ 3-5=0
5. (X -5) mũ 2016 = (X-5) mũ 2018
6. (3X -2) mũ 20= (3X-1) mũ 20
7. (3X -1) mũ 10 = (3X-1) mũ 20
8. (2X -1) mũ 50 = 2X-1
9. (X phần 3 -5) mũ 2000= ( X phần 3-5) mũ 2008
1. \(6x^3-8=40\\ 6x^3=48\\ x^3=8\\ \Rightarrow x=2\)Vậy x = 2
2. \(4x^5+15=47\\ 4x^5=32\\ x^5=8\\ \Rightarrow x\in\varnothing\left(\text{vì }x\in N\right)\)Vậy x ∈ ∅
3. \(2x^3-4=12\\ 2x^3=16\\ x^3=8\\ \Rightarrow x=2\)Vậy x = 2
4. \(5x^3-5=0\\ 5x^3=5\\ x^3=1\\ \Rightarrow x=1\)Vậy x = 1
5. \(\left(x-5\right)^{2016}=\left(x-5\right)^{2018}\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\x-5=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)Vậy \(x\in\left\{5;6\right\}\)
6. \(\left(3x-2\right)^{20}=\left(3x-1\right)^{20}\\ \Rightarrow3x-2=3x-1\\ 3x-3x=2-1\\ 0=1\left(\text{vô lí}\right)\)Vậy x ∈ ∅
7. \(\left(3x-1\right)^{10}=\left(3x-1\right)^{20}\\ \left(3x-1\right)^{10}=\left[\left(3x-1\right)^2\right]^{10}\\ \Rightarrow\left(3x-1\right)^2=3x-1\\ \left(3x-1\right)^2-\left(3x-1\right)=0\\ \left(3x-1\right)\left[\left(3x-1\right)-1\right]=0\\ \left(3x-1\right)\left(3x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x-1=0\\3x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x=1\\3x=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{1}{3}\left(\text{loại vì }x\in N\right)\\x=\frac{2}{3}\left(\text{loại vì }x\in N\right)\end{matrix}\right.\)Vậy x ∈ ∅
8. \(\left(2x-1\right)^{50}=2x-1\\ \left(2x-1\right)^{50}-\left(2x-1\right)=0\\ \left(2x-1\right)\left[\left(2x-1\right)^{49}-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}2x-1=0\\\left(2x-1\right)^{49}=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=1\\2x-1=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\frac{1}{2}\left(\text{loại vì }x\in N\right)\\x=1\left(t/m\right)\end{matrix}\right.\)Vậy x = 1
9. \(\left(\frac{x}{3}-5\right)^{2000}=\left(\frac{x}{3}-5\right)^{2008}\\ \left(\frac{x}{3}-5\right)^{2008}-\left(\frac{x}{3}-5\right)^{2000}=0\\ \left(\frac{x}{3}-5\right)^{2000}\left[\left(\frac{x}{3}-5\right)^8-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}\left(\frac{x}{3}-5\right)^{2000}=0\\\left(\frac{x}{3}-5\right)^8=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\frac{x}{3}-5=0\\\frac{x}{3}-5=1\\\frac{x}{3}-5=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\frac{x}{3}=5\\\frac{x}{3}=6\\\frac{x}{3}=4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\cdot3=15\\x=6\cdot3=18\\x=4\cdot3=12\end{matrix}\right.\)Vậy \(x\in\left\{15;18;12\right\}\)
\(1.6x^3-8=40\\ \Leftrightarrow6x^3=48\\ \Leftrightarrow x^3=8\Leftrightarrow x^3=2^3=\left(-2\right)^3\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy \(x\in\left\{2;-2\right\}\)
\(2.4x^3+15=47\) (T nghĩ đề là mũ 3)
\(\Leftrightarrow4x^3=32\Leftrightarrow x^3=8=2^3=\left(-2\right)^3\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy \(x\in\left\{2;-2\right\}\)
Câu 3, 4 tương tự nhé.
\(5.\left(x-5\right)^{2016}=\left(x-5\right)^{2018}\\ \Leftrightarrow\left(x-5\right)^{2018}-\left(x-5\right)^{2016}=0\\ \Leftrightarrow\left(x-5\right)^{2016}\left[\left(x-5\right)^2-1\right]=0\\ \Leftrightarrow\left(x-5\right)^{2016}\left(x-5-1\right)\left(x-5+1\right)=0\\ \Leftrightarrow\left(x-5\right)^{2016}\left(x-6\right)\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-5\right)^{2016}=0\\x-6=0\\x-4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x-5=0\\x=6\\x=4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=6\\x=4\end{matrix}\right.\)
Vậy \(x\in\left\{4;5;6\right\}\)
Bài 2: Tìm số tự nhiên x,biết:
1, 2 mũ x = 4
2, 2 mũ x = 8
3, 2 mũ x = 16
4, 2 mũ x = 1
5, 3 mũ x = 9
6, 3 mũ x = 81
7, 3 mũ x = 27
8, 5 mũ x = 25
9, 5 mũ x = 125
10, 8 mũ x = 64
11, 3 mũ x + 1 = 3 mũ 2
12, 2 mũ 2 x + 1 = 2 mũ 7
13, 5 mũ x - 1 = 5 mũ 2
14, 5 mũ 2 x - 4 = 5 mũ 10
15, 6 x + 4 = 6 mũ 10
16, 2 mũ 2 x - 3 = 2 mũ 9
17, 7 mũ 2 x - 3 = 7 mũ 7
18, 8 mũ x - 2 = 1
19, 9 mũ x - 8 = 81
20, 10 mũ 2 x - 1 = 1000
Giúp mình với,mình đang cần gấp !!
Bài 1:
2\(x\) = 4
2\(^x\) = 22
\(x=2\)
Vậy \(x=2\)
Bài 2:
2\(^x\) = 8
2\(^x\) = 23
\(x=3\)
Vậy \(x=3\)
Bài 3
2\(^x\) = 16
2\(^x\) = 24
\(x=4\)
Vậy \(x=4\)