Bài 1 : 

Ta có : 

\(A=\frac{\frac{3}{4}-\frac{3}{11}+\frac{3}{13}}{\frac{5}{7}-\frac{5}{11}+\frac{5}{13}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{4}-\frac{5}{6}+\frac{5}{8}}\)

\(A=\frac{3\left(\frac{1}{4}-\frac{1}{11}+\frac{1}{13}\right)}{5\left(\frac{1}{7}-\frac{1}{11}+\frac{1}{13}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{2}\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}\right)}\)

\(A=\frac{3}{5}+\frac{1}{\frac{5}{2}}\)

\(A=\frac{3}{5}+\frac{2}{5}\)

\(A=1\)

\(b)\) Áp dụng tính chất dãy tỉ số bằng nhau ta có : 

\(\frac{y+z-x}{x}=\frac{z+x-y}{y}=\frac{x+y-z}{z}=\frac{y+z-x+z+x-y+x+y-z}{x+y+z}=\frac{2\left(x+y+z\right)}{x+y+z}=2\)

Đo đó : 

\(\frac{y+z-x}{x}=2\)\(\Rightarrow\)\(y+z=3x\)\(\left(1\right)\)

\(\frac{z+x-y}{y}=2\)\(\Rightarrow\)\(x+z=3y\)\(\left(2\right)\)

\(\frac{x+y-z}{z}=2\)\(\Rightarrow\)\(x+y=3z\)\(\left(3\right)\)

Lại có : \(B=\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)=\frac{x+y}{y}.\frac{y+z}{z}.\frac{x+z}{x}\)

Thay (1), (2) và (3) vào \(B=\frac{x+y}{y}.\frac{y+z}{z}.\frac{x+z}{x}\) ta được : 

\(B=\frac{2z}{y}.\frac{2x}{z}.\frac{2y}{x}=\frac{8xyz}{xyz}=8\)

Vậy \(B=8\)

Chúc bạn học tốt ~ 

bạn phùng minh quân câu 1 a tại sao lại rút gọn được \(\frac{3.\left(\frac{1}{4}-\frac{1}{11}+\frac{1}{13}\right)}{5\left(\frac{1}{7}-\frac{1}{11}+\frac{1}{13}\right)}=\frac{3}{5}\) vậy nó không cùng nhân tử mà 

câu b \(\frac{y+z-x+z+x-y+x+y-z}{x+y+z}=\frac{\left(y-y+y\right)+\left(-x+x+x\right)+\left(z+z-z\right)}{x+y+z}=\frac{x+y+z}{x+y+z}=1\)sao lại ra bằng 2

(mình chỉ góp ý thôi nha tại mình làm thấy nó sai sai) 

A=1

B=8

ĐÚNG K Z 

=)))))

\(\frac{4}{\frac{2}{5}}:\left(-\frac{33}{10}\right)+x=-\frac{1}{\frac{5}{6}}\)

\(10:\left(-\frac{33}{10}\right)+x=-\frac{6}{5}\)

\(-\frac{100}{33}+x=-\frac{6}{5}\)

\(x=\frac{302}{165}\)

a: \(\left(x+\dfrac{1}{4}\right)+\left(3x-4\right)+2\left(x-3\right)=1\)

=>\(x+\dfrac{1}{4}+3x-4+2x-6=1\)

=>\(6x-\dfrac{39}{4}=1\)

=>\(6x=1+\dfrac{39}{4}=\dfrac{43}{4}\)

=>\(x=\dfrac{43}{4}:6=\dfrac{43}{24}\)

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

=>\(2x-6=3x+6-x+1\)

=>2x-6=2x+7

=>-6=7(vô lý)

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

=>\(x^2+3x+x^2-2x=2x^2-2x\)

=>3x-2x=-2x

=>3x=0

=>x=0

d: \(\left(x-1\right)\cdot3x-2\left(x+2\right)-2x=x\left(x-1\right)\)

=>\(3x^2-3x-2x-4-2x=x^2-x\)

=>\(3x^2-7x-4-x^2+x=0\)

=>\(2x^2-6x-4=0\)

=>\(x^2-3x-2=0\)

=>\(x=\dfrac{3\pm\sqrt{17}}{2}\)

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

\(2\left(x^2+3x-2x-6\right)-x^2+4=0\)

\(2x^2+6x-4x-12-x^2+4=0\)

\(x^2+2x-8=0\)

\(x^2+4x-2x-8=0\)

\(x\left(x+4\right)-2\left(x+4\right)=0\)

\(\left(x+4\right)\left(x-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+4=0\rightarrow x=\left(-4\right)\\x-2=0\rightarrow x=2\end{cases}}\)

3/ 

a/ \(2\left(x+1\right)^2-3\left(x-1\right)^2+\left(x+2\right)\left(5-x\right)\)

\(=2\left(x^2+2x+1\right)-3\left(x^2-2x+1\right)+\left(5x-x^2+10-2x\right)\)

\(=2x^2+4x+2-3x^2+6x-3+5x-x^2+10-2x\)

\(=-2x^2+13x+9\)

b/ \(\left(3x-1\right)^3+\left(3x-1\right)^3-6x^2+9\)

\(=2\left(3x-1\right)^3-6x^2+9\)

\(=2\left(\left(3x\right)^3-3\left(3x\right)^2\cdot1+3\cdot3x\cdot1-1\right)-6x^2+9\)

\(=2\left(27x^3-27x^2+9x-1\right)-6x^2+9\)

\(=54x^3-54x^2+18x-2-6x^2+9\)

\(=54x^3-60x^2+18x+7\)

Số hơi dài, nên dễ tính sai -,- tính mik hay cẩu thả có j sai ibbb ạ

2) 2.(x - 2).(x + 3) - x² + 4 = 0

<=> x² + 2x - 8 = 0

<=> (x - 2).(x + 4) = 0

        x - 2 = 0 hoặc x + 4 = 0

        x = 0 + 2         x = 0 - 4

        x = 2               x = -4

=> x = 2 hoặc x = -4

3) a) 2.(x + 1)² - 3.(x - 1)² + (x + 2).(5 - x)

= 2.(x² + 2x + 1) - 3.(x² - 2x + 1) + (x + 2).(5 - x)

= 2x² + 4x + 2 - 3x² + 6x - 3 + (x + 2).(5 - x)

= 2x² + 4x + 2 - 3x² + 6x - 3 + 3x - x² + 10

= (2x² - 3x² - x²) + (4x + 6x + 3x) + (2 - 3 + 10)

= -2x² + 13x + 9

b) (3x - 1)³ + (3x - 1)³ - 6x² + 9

= 2.(3x - 1)³ - 6x² + 9

= 2.(27x³ - 27x² + 9x - 1) - 6x² + 9

= 54x³ - 54x² + 18x - 2 - 6x² + 9

= 54x³ + (-54x² - 6x²) + 18x + (-2 + 9)

= 54x³ - 60x + 18x + 7

4) a) A = (x - 5).(x + 2) + 3.(x - 2).(x + 2) - (3x - 1)² + 5x²

A = (x - 5).(x + 2) + 3.(x - 2).(x + 3) - (9x² - 6x + 1) + 5x²

A = x² - 3x - 10 + 3x² - 12 - (9x² - 6x + 1) + 5x²

A = x² - 3x - 10 + 3x² - 12 - 9x² + 6x - 1 + 5x²

A = (x² + 3x² - 9x² + 5x²) + (-3x + 6x) + (-10 - 12 - 1)

A = 3x - 23 (1)

b) Thay x = 1/2 vào (1), ta có:

A = 3x - 23 = 3.(1/2) - 23

                   = 3/2 - 23

                   = -43/2

A khi x = 1/2 là -43/2

1)x=1,5

2)-5

3)14

4)0 có cap a;b thoa man de bai(điền số 0 vào)

5)-2011,đúng rồi đấy

6)Pmin=3,7

tick nhé,tớ thi violymic rồi

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