thôi đi Đỗ Thùy Dương

làm j chửi lăm thế, k trả lời thì thôi

Đúng đó Vũ Diệu Ngân Hà

a,676 phần 9

b,Ko bt,sao lại có y?

Bạn thu gọn các đa thức rồi thay thế vào sẽ tính ra ngay nha!

a) 25x² - 10xy + y²

= (5x)² - 2.5x.y + y²

= (5x - y)²

b) 4/9 x² + 20/3 xy + + 25y²

= (2/3 x)² + 2.2/3 x.5y + (5y)²

= (2/3 x + 5y)²

c) 9x² - 12x + 4

= (3x)² - 2.3x.2 + 2²

= (3x - 2)²

d) Sửa đề: 16u²v⁴ - 8uv² + 1

= (4uv²)² - 2.4uv².1 + 1²

= (4uv² - 1)²

chỉ cần giải mỗi c và d thui

giúp với sos

\(25x^2-10xy+y^2=\left(5x\right)^2-2.5x.y+y^2=\left(5x-y\right)^2\)

\(\dfrac{4}{9}x^2+\dfrac{20}{3}xy+25y^2=\left(\dfrac{2}{3}x\right)^2+2.\dfrac{2}{3}x.5y+\left(5y\right)^2=\left(\dfrac{2}{3}x+5y\right)^2\)

\(a)\)

\(21\left(x+3\right)^3:\left(3x+9\right)^2\)

\(=[21\left(x+3\right)^3]:[3^2\left(x+3\right)^2]\)

\(=7\left(x+3\right):3\)

Thay vào ta được: \(7.\frac{\left(-6+3\right)}{3}=7.\left(-3\right):3=-7\)

\(b)\)

Thay vào ta được:

\(\left(2.2^2-5.2+3\right)^4:[\left(2.2-3\right)^3:\left(2-1\right)^2]\)

\(=\left(2.4-10+3\right)^4:[\left(4-3\right)^31^2]\)

\(=1^4:\left(1^3.1\right)\)

\(=1:1\)

\(=1\)

\(c)\)

Thay vào ta được:

\(36.10^4.7^3:\left(-6.10^3.7^2\right)\)

\(=-6.10.7\)

\(=-420\)

a, \(5x^2-10xy+5y^2=5\left(x^2-2xy+y^2\right)=5.\left(x-y\right)^2\)

b, \(x^2-4x+4-y^2=\left(x^2-4x+4\right)-y^2=\left(x-2\right)^2-y^2\)

\(=\left(x-2-y\right)\left(x-2+y\right)\)

c, \(3x^2-2x-5=3x^2-5x+3x-5=x\left(3x-5\right)+3x-5\)

\(=\left(3x-5\right)\left(x+1\right)\)

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 \(x\in\left\{4;5;6\right\}\)