a) \(\frac{1}{m+1}+\frac{1}{\left(m+1\right)\left(2m+1\right)}\)

\(=\frac{2m+1}{\left(m+1\right)\left(2m+1\right)}+\frac{1}{\left(m+1\right)\left(2m+1\right)}\)

\(=\frac{2m+2}{\left(m+1\right)\left(2m+1\right)}\)

\(=\frac{2\left(m+1\right)}{\left(m+1\right)\left(2m+1\right)}\)

\(=\frac{2}{2m+1}=\frac{4}{4m+2}\left(đpcm\right)\)

b) \(\frac{1}{m+2}+\frac{1}{\left(m+1\right)\left(m+2\right)}+\frac{1}{\left(m+1\right)\left(4m+3\right)}\)

\(=\frac{m+1}{\left(m+1\right)\left(m+2\right)}+\frac{1}{\left(m+1\right)\left(m+2\right)}+\frac{1}{\left(m+1\right)\left(4m+3\right)}\)

\(=\frac{m+2}{\left(m+1\right)\left(m+2\right)}+\frac{1}{\left(m+1\right)\left(4m+3\right)}\)

\(=\frac{1}{m+1}+\frac{1}{\left(m+1\right)\left(4m+3\right)}\)

\(=\frac{4m+3}{\left(m+1\right)\left(4m+3\right)}+\frac{1}{\left(m+1\right)\left(4m+3\right)}\)

\(=\frac{4m+4}{\left(m+1\right)\left(4m+3\right)}\)

\(=\frac{4\left(m+1\right)}{\left(m+1\right)\left(4m+3\right)}\)

\(=\frac{4}{4m+3}\left(đpcm\right)\)

Đổi \(15m/s=54km/h\)

Thời gian ô tô đi nửa đoạn đầu:

\(t_1=\frac{S_1}{v_1}=\frac{\frac{S}{2}}{54}=\frac{S}{108}\left(h\right)\)

Mặt khác,ta có:

\(S_2+S_3=\frac{S}{2}\)

\(\Leftrightarrow v_2\cdot t_2+v_3\cdot t_3=\frac{S}{2}\)

\(\Leftrightarrow v_2\cdot\frac{t'}{2}+v_3\cdot\frac{t'}{2}=\frac{S}{2}\)

\(\Leftrightarrow45t'+15t'=S\)

\(\Rightarrow t'=\frac{S}{60}\)

Ta có:

\(v_{tb}=\frac{S_1+S_2+S_3}{t_1+t_2+t_3}=\frac{S}{\frac{S}{108}+\frac{S}{60}}=\frac{270}{7}km/h\)

Đúng không ta ??

Câu hỏi của Khoa Nguyễn Đăng - Toán lớp 8 - Học toán với OnlineMath

Em tham khảo nhé!

\(a+b+c\Rightarrow\left(a+b+c\right)^2=0\)

\(\Leftrightarrow a^2+b^2+c^2+2ab+2ac+2bc=0\)

\(\Leftrightarrow ab+bc+ac=\frac{-a^2-b^2-c^2}{2}\)

\(\Rightarrow2\left(ab+bc+ac\right)^2=\frac{\left(a^2+b^2+c^2\right)^2}{2}\)(1)

Lại có : \(\left(a^2+b^2+c^2\right)^2=a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+a^2c^2\right)\)

                                               \(=a^4+b^4+c^4+2\left(ab+bc+ac\right)^2-2abc\left(a+b+c\right)\)

                                               \(=a^4+b^4+c^4+2\left(ab+bc+ac\right)^2\)( do a + b + c = 0 )

Thay vào ( 1 )

\(2\left(ab+bc+ca\right)^2=\frac{a^4+b^4+c^4}{2}+\left(ab+cb+ac\right)^2\)

\(\Rightarrow\left(ab+bc+ca\right)^2=\frac{a^4+b^4+c^4}{2}\)

\(\Rightarrowđpcm\)

Ta có:
\(\left(a+b+c\right)^2=3\left(ab+bc+ca\right)\)

\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)-3\left(ab+bc+ca\right)=0\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2\left(ab+bc+ca\right)=0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

\(\Leftrightarrow a=b=c\) hay tam giác ABC đều.

a) \(x^2-4x-7=0\)

Ta có: \(\Delta=4^2+4.28=128,\sqrt{\Delta}=\sqrt{128}\)

pt có 2 nghiệm:

\(x_1=\frac{4+\sqrt{128}}{2}\);\(x_2=\frac{4-\sqrt{128}}{2}\)

b) \(x^2-x-11=0\)

Ta có: \(\Delta=1^2+4.11=45,\sqrt{\Delta}=\sqrt{45}\)

pt có 2 nghiệm:

\(x_1=\frac{1+\sqrt{45}}{2}\)\(x_2=\frac{1-\sqrt{45}}{2}\)

Diệu Linh_face

\(6x^2+x-2\)

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

\(=3x\left(2x-1\right)+2\left(2x-1\right)\)

\(=\left(2x-1\right)\left(3x+2\right)\)

thak nha

Mik đang cần gấp mn ơi!!!

\(\frac{a}{x+1}+\frac{b}{x-1}=\frac{5x+1}{x^2-1}\)

\(\Leftrightarrow\frac{a\left(x-1\right)+b\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}=\frac{5x+1}{\left(x+1\right)\left(x-1\right)}\)

\(\Leftrightarrow ax-a+bx+b=5x+1\)

\(\Leftrightarrow x\left(a+b\right)-a+b=5x+1\)

\(\Rightarrow\hept{\begin{cases}a+b=5\\b-a=1\end{cases}\Rightarrow\hept{\begin{cases}a=2\\b=3\end{cases}}}\)

a) \(2x^2+3x-8=0\)

Ta có: \(\Delta=3^2+4.2.8=73\)

pt có 2 nghiệm

\(x_1=\frac{-3+\sqrt{73}}{4}\);\(x_1=\frac{-3-\sqrt{73}}{4}\)

d) \(\left(x^2+2x\right)^2-2\left(x^2+2x\right)-3=0\)

Đặt \(x^2+2x=t\)

\(pt\Leftrightarrow t^2-2t-3=0\)

Ta có: \(\Delta=2^2+4.3=16,\sqrt{\Delta}=4\)

pt trên có 2 nghiệm

\(x_1=\frac{2+4}{2}=3;x_2=\frac{2-4}{2}=-1\)

\(\Rightarrow\orbr{\begin{cases}x^2+2x=3\\x^2+2x=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}\left(x+3\right)\left(x-1\right)=0\\\left(x+1\right)^2=0\end{cases}}\)

\(\Rightarrow x\in\left\{-3;-1;1\right\}\)

c) \(x^4+8x^3+19x^2+12x=0\)

\(\Leftrightarrow x^4+4x^3+4x^3+16x^2+3x^2+12x=0\)

\(\Leftrightarrow\left(x^4+4x^3+3x^2\right)+\left(4x^3+16x^2+12x\right)=0\)

\(\Leftrightarrow x\left(x^3+4x^2+3x\right)+4\left(x^3+4x^2+3x\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x^3+4x^2+3x\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x^3+x^2+3x^2+3x\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left[x^2\left(x+1\right)+3x\left(x+1\right)\right]=0\)

\(\Leftrightarrow\left(x+4\right)\left(x^2+3x\right)\left(x+1\right)=0\)

\(\Leftrightarrow x\left(x+1\right)\left(x+3\right)\left(x+4\right)=0\)

\(\Leftrightarrow x\in\left\{0;-1;-3;-4\right\}\)

Ta có:

a) 6x²y - 3y² - 2x² + y = (6x²y - 2x²) - (3y² - y) = 2x²(3y - 1) - y(3y - 1) = (2x² - y)(3y - 1)

b)  2x² + x - 4xy - 2y + 2x + 1 = (x² + x) - (4xy + 2y) + (x² + 2x + 1) = x(x + 1) - 2y(2x + 1) + (x + 1)²

 = (x + x + 1)(x + 1) - 2y(2x + 1) = (2x + 1)(x + 1) - 2y(2x + 1) = (2x + 1)(x + 1 - 2y)

c) 16x²y - 4xy² - 4x³ + x²y = 4xy(4x - y) - x²(4x - y) = (4xy - x²)(4x - y)

d) 4x² - 20x + 25 - 36y² = (2x  - 5)² - (6y)² = (2x - 5 - 6y)(2x  - 5 + 6y)

e) x² - 4y² + 6x - 4y + 8 = (x² + 6x + 9) - (4y² + 4y + 1) = (x + 3)² - (2y + 1)² = (x + 3 - 2y - 1)(x + 3 + 2y + 1) = (x + 2 - 2y)(x + 4 + 2y)

g) Ta có : x¹⁰ + x⁵ + 1

= (x¹⁰ - x) + (x⁵ - x²) + (x² + x + 1)

= x(x⁹ - 1) + x²(x³ - 1) + (x² + x + 1)

= x(x³ - 1)(x⁶ + x³ + 1) + x²(x³ - 1) + (x² + x + 1)

= (x⁷ + x⁴ + x)(x - 1)(x² + x + 1) + x²(x - 1)(x² + x + 1) + (x² + x + 1)

= (x² + x + 1)(x⁸ - x⁷ + x ⁵ - x⁴ + x² - x + x⁴ + x³ + x² + 1)

= (x² + x + 1)(x⁸ - x⁷ + x⁵ + x³ - x + 1)

h) TT trên (dài dòng ktl)