Đề sai vì khi lấy tam giác vuông có độ dài hai cạnh góc vuông là 3 và 4, cạnh huyền là 5 thì chiều cao là:

3 . 4 : 5 = 2,4

Mà 2,4 + 5 > 3 + 4 (vì 7,4 > 7)

 Theo mình thì nó phải là ngược lại mới đúng: Tổng cạnh huyền và đường cao tương ứng luôn lớn hơn tổng hai cạnh góc vuông. (*)

 Chứng minh:

 Ta có \(AB^2+AC^2=BC^2\) (định lý Pythagoras)

\(\Leftrightarrow AB^2+2AB.AC+BC^2=BC^2+2AH.BC\)

\(\Leftrightarrow\left(AB+AC\right)^2=BC\left(2AH+BC\right)\)  

 Mà \(BC\left(2AH+BC\right)\le\left(\dfrac{BC+2AH+BC}{2}\right)^2\) \(=\left(AH+BC\right)^2\) (áp dụng bất đẳng thức  \(ab\le\left(\dfrac{a+b}{2}\right)^2\))

 Dấu "=" không thể xảy ra vì khi đó \(BC=BC+2AH\), vô lí.

 Vậy \(\left(AB+AC\right)^2=BC\left(2AH+BC\right)< \left(AH+BC\right)^2\)

 \(\Leftrightarrow AB+AC< AH+BC\). 

 Vậy (*) được chứng minh.

Ta biến đổi \(1+\dfrac{1}{\left(k-1\right)\left(k+1\right)}\) 

\(=\dfrac{\left(k-1\right)\left(k+1\right)+1}{\left(k-1\right)\left(k+1\right)}\)

\(=\dfrac{\left(k^2-1\right)+1}{\left(k-1\right)\left(k+1\right)}\)

\(=\dfrac{k^2}{\left(k-1\right)\left(k+1\right)}\) (với \(k\ge2\))

Do đó \(P=\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)...\left(1+\dfrac{1}{2020.2022}\right)\left(1+\dfrac{1}{2021.2023}\right)\)
\(P=\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}...\dfrac{2021^2}{2020.2022}.\dfrac{2022^2}{2021.2023}\)

\(P=\dfrac{2.2022}{2023}\)

\(P=\dfrac{4044}{2023}\)

5x² - 16x = 0

x(5x - 16) = 0

x = 0 hoặc 5x - 16 = 0

*) 5x - 16 = 0

5x = 16

x = 16/5

Vậy x = 0; x = 16/5

Sắp xếp:

A(x) = -2x² + 3x - 4x³ + 3/5 - 5x⁴

= -5x⁴ - 4x³ - 2x² + 3x + 3/5

B(x) = 3x⁴ + 1/5 - 7x² + 5x³ - 9x

= 3x⁴ + 5x³ - 7x² - 9x + 1/5

--------

A(x) + B(x) = (-5x⁴ - 4x³ - 2x² + 3x + 3/5) + (3x⁴ + 5x³ - 7x² - 9x + 1/5)

= -5x⁴ - 4x³ - 2x² + 3x + 3/5 + 3x⁴ + 5x³ - 7x² - 9x + 1/5

= (-5x⁴ + 3x⁴) + (-4x³ + 5x³) + (-2x² - 7x²) + (3x - 9x) + (3/5 + 1/5)

= -2x⁴ + x³ - 9x² - 6x + 4/5

-----------

A(x) - B(x) = (-5x⁴ - 4x³ - 2x² + 3x + 3/5) - (3x⁴ + 5x³ - 7x² - 9x + 1/5)

= -5x⁴ - 4x³ - 2x² + 3x + 3/5 - 3x⁴ - 5x³ + 7x² + 9x - 1/5

= (-5x⁴ - 3x⁴) + (-4x³ - 5x³) + (-2x² + 7x²) + (3x + 9x) + (3/5 - 1/5)

= -8x⁴ - 9x³ + 5x² + 12x + 2/5

Sắp xếp:

\(A=-2x^2+3x-4x^3+\dfrac{3}{5}-5x^4=-5x^4-4x^3-2x^2+3x+\dfrac{3}{5}\)

\(B=3x^4+\dfrac{1}{5}-7x^2+5x^3-9x=3x^4+5x^3-7x^2-9x+\dfrac{1}{5}\) 

Tính: 

\(A\left(x\right)+B\left(x\right)\)

\(=-5x^4-4x^3-2x^2+3x+\dfrac{3}{5}+3x^4+5x^3-7x^2-9x+\dfrac{1}{5}\)

\(=\left(-5x^4+3x^4\right)+\left(-4x^3+5x^3\right)+\left(-2x^2-7x^2\right)+\left(3x-9x\right)+\left(\dfrac{3}{5}+\dfrac{1}{5}\right)\)

\(=-2x^4+x^3-9x^2-6x+\dfrac{4}{5}\)

\(A\left(x\right)-B\left(x\right)\)

\(=\left(-5x^4-4x^3-2x^2+3x+\dfrac{3}{5}\right)-\left(3x^4+5x^3-7x^2-9x+\dfrac{1}{5}\right)\)

\(=-5x^4-4x^3-2x^2+3x+\dfrac{3}{5}-3x^4-5x^3+7x^2+9x-\dfrac{1}{5}\)

\(=\left(-5x^4-3x^4\right)+\left(-4x^3-5x^3\right)+\left(-2x^2+7x^2\right)+\left(3x+9x\right)+\left(\dfrac{3}{5}-\dfrac{1}{5}\right)\)

\(=-8x^4-8x^3+5x^2+12x+\dfrac{2}{5}\)

cần nhanh

giúp với

\(\dfrac{x-3}{x+7}\) = \(\dfrac{1}{6}\)

6.(\(x-3\)) = \(x+7\)

6\(x-18\) = \(x+7\)

6\(x\) - \(x\) = 18 + 7

5\(x\) = 25

\(x\) = 5

Thay \(x=5\) vào \(\dfrac{x+4}{9}=\dfrac{y}{13}\) ta có:

                          \(\dfrac{5+4}{9}\) = \(\dfrac{y}{13}\)

                            1 = \(\dfrac{y}{13}\)

                            y = 13

Vậy (\(x;y\)) =(5; 13)

Khó nha 

a) \(A\left(x\right)+B\left(x\right)\)

\(=\left(-x^6+x^4-4x^3+x^2-5\right)+\left(2x^5-x^4-x^3+x^2+x-1\right)\)

\(=-x^6+x^4-4x^3+x^2-5+2x^5-x^4-x^3+x^2+x-1\)

\(=-x^6+2x^5-5x^3+2x^2+x-6\)

b) \(A\left(x\right)-B\left(x\right)\)

\(=\left(-x^6+x^4-4x^3+x^2-5\right)-\left(2x^5-x^4-x^3+x^2+x-1\right)\)

\(=-x^6+x^4-4x^3+x^2-5-2x^5+x^4+x^3-x^2-x+1\)

\(=-x^6-2x^5+2x^4-3x^3-x-4\)

Ta có: \(A\left(x\right)=-x^6+x^4-4x^3+x^2-5\)

và \(B\left(x\right)=2x^5-x^4-x^3+x^2+x-1\)

a) \(A\left(x\right)+B\left(x\right)=\left(-x^6+x^4-4x^3+x^2-5\right)+\left(2x^5-x^4-x^3+x^2+x-1\right)\)

\(=-x^6+2x^5+\left(x^4-x^4\right)+\left(-4x^3-x^3\right)+\left(x^2+x^2\right)+x+\left(-5-1\right)\)

\(=-x^6+2x^5-5x^3+2x^2+x-6\)

b) \(A\left(x\right)-B\left(x\right)=\left(-x^6+x^4-4x^3+x^2-5\right)-\left(2x^5-x^4-x^3+x^2+x-1\right)\)

\(=-x^6+x^4-4x^3+x^2-5-2x^5+x^4+x^3-x^2-x+1\)

\(=-x^6-2x^5+\left(x^4+x^4\right)+\left(-4x^3+x^3\right)+\left(x^2-x^2\right)-x+\left(-5+1\right)\)

\(=-x^6-2x^5+2x^4-3x^3-x-4\)

\(\left(x-1\right)^2=\left(x-1\right)^4\)

\(\Rightarrow\left(x-1\right)^4-\left(x-1\right)^2=0\)

\(\Rightarrow\left(x-1\right)^2\left[\left(x-1\right)^2-1\right]=0\) 

+) \(\left(x-1\right)^2=0\)

\(\Rightarrow x-1=0\)

\(\Rightarrow x=1\)

+) \(\left(x-1\right)^2-1=0\)

\(\Rightarrow\left(x-1\right)^2=1\)

\(\Rightarrow\left(x-1\right)^2=1^2\)

TH1: \(x-1=1\Rightarrow x=1+1=2\)

TH2: \(x-1=-1\Rightarrow x=-1+1=0\)

Vậy: \(x\in\left\{1;2;0\right\}\)

\(\left(x-1\right)^2=\left(x-1\right)^4\)

\(\Rightarrow\left(x-1\right)^2-\left(x-1\right)^4=0\)

\(\Rightarrow\left(x-1\right)^2\left[1-\left(x-1\right)^2\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\1-\left(x-1\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-1=0\\\left(x-1\right)^2=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x-1=1\\x-1=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=2\\x=0\end{matrix}\right.\)

Vậy \(x\in\left\{1;2;0\right\}\)

for you