Bài 1: 

\(B=\frac{\frac{1}{2}+\frac{3}{4}-\frac{5}{6}}{\frac{1}{4}+\frac{3}{8}-\frac{5}{12}}+\frac{\frac{3}{4}+\frac{3}{5}-\frac{3}{8}}{\frac{1}{4}+\frac{1}{5}-\frac{1}{8}}\)\(=\frac{\frac{1}{2}+\frac{3}{4}-\frac{5}{6}}{\frac{1}{2}\left(\frac{1}{2}+\frac{3}{4}-\frac{5}{6}\right)}+\frac{3\left(\frac{1}{4}+\frac{1}{5}-\frac{1}{8}\right)}{\frac{1}{4}+\frac{1}{5}-\frac{1}{8}}\) 

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

                                                  Vậy B=5

Bài 2:

a) x³ - 36x = 0  

=>  x(x²-36)=0

=>  x(x²+6x-6x-36)=0 

=> x[x(x+6)-6(x+6) ]=0

=> x(x+6)(x-6)=0

\(\Rightarrow\orbr{\begin{cases}^{x=0}x+6=0\\x-6=0\end{cases}}\)

 \(\Rightarrow\orbr{\begin{cases}^{x=0}x=-6\\x=6\end{cases}}\)

                                  Vậy x=0; x=-6; x=6

b)  (x - y = 4 => x=4+y)

 x−3y−2 =32  

=>2(x-3) = 3(y-2)

=>2x-6= 3y-6

=>2x-3y=0

=>2(4+y)-3y=0

=>8+2y-3y=0

=>8-y=0

=>y=8 (thỏa mãn)

Do đó x=4+y=4+8=12 (thỏa mãn)

         Vậy x=12 và y =8

B= 1/2 + 3/4 - 5/6/1/2(1.2 + 3/4 - 5/6) + 3(1/4+ 1/5 - 1/8)/ 1/4  1/5 - 1/8 

B= 1/ 1/2 + 3

B= 2+3

B=5

B2:

a) x^3 - 36x = 0

x(x^2 - 36) = 0

=> x=0  hoặc x^2-36=0

=> x= 0 hoặc x^2=36

=> x=0 hoặc x= +- 6

b) x-y = 4 => x= 4+y

thay x=4+y vào x- 3/ y-2=3/2, có:

4+y-3/ y+2 = 3/2

y+1/ y+2 = 3/2

y+2 -1/ y+2 = 3/2

1 - 1/y+2 = 3/2

1/y+2= 1-3/2

1/y+2 = -1/2

=> y+2 = -2

=> y= -4

Dp x= 4+y => x= 4-4

=> x=0

Vậy x=0 và y=-4

1 

Ez lắm =)

Bài 1:

Với mọi gt \(x,y\in Q\) ta luôn có: 

\(x\le\left|x\right|\) và \(-x\le\left|x\right|\) 

\(y\le\left|y\right|\) và \(-y\le\left|y\right|\Rightarrow x+y\le\left|x\right|+\left|y\right|\) và \(-x-y\le\left|x\right|+\left|y\right|\)

Hay: \(x+y\ge-\left(\left|x\right|+\left|y\right|\right)\)

Do đó: \(-\left(\left|x\right|+\left|y\right|\right)\le x+y\le\left|x\right|+\left|y\right|\)

Vậy: \(\left|x+y\right|\le\left|x\right|+\left|y\right|\)

Dấu "=" xảy ra khi: \(xy\ge0\)

Bài 3: 

Ta có: \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=\frac{x+y+z}{a+b+c}=x+y+z\) (vì a + b + c = 1)

Do đó: \(\left(x+y+z\right)^2=\frac{x^2}{a^2}=\frac{y^2}{b^2}=\frac{z^2}{c^2}=\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=x^2+y^2+z^2\) (vì a² + b² + c² = 1)

Vậy: (x + y + z)² = x² + y² + z²

a) Ta có : \(\frac{x}{3}-\frac{4}{y}=\frac{1}{5}\)

\(\Rightarrow\frac{x}{3}-\frac{1}{5}=\frac{4}{y}\)

\(\Rightarrow\frac{x.5}{15}-\frac{3}{15}=\frac{4}{y}\)

\(\Rightarrow\frac{x.5-3}{15}=\frac{4}{y}\)

\(\Rightarrow\left(x.5-3\right).y=15.4\)

\(\Rightarrow x.5.y-3.5=60\)

\(\Rightarrow xy5-15=60\)

 \(\Rightarrow xy5=60+15\)

\(\Rightarrow xy5=75\) 

\(\Rightarrow xy=75\div5\)

\(\Rightarrow xy=15\)

\(\Rightarrow xy=1.15=3.5=\left(-15\right)\left(-1\right)=\left(-3\right)\left(-5\right)=\left(-5\right)\left(-3\right)=\left(-1\right)\left(-15\right)=5.3=15.1\)

Do đó x = 1 thì y = 15

x = 3 thì y =5

x = -15 thì y = -1

x = -3 thì y = -5

x = -5 thì y = -3

x = -1 thì y = -15

x = 5 thì y = 3

x = 15 thì y = 1

Câu b) tạm thời ko bít làm =.= 

Bài 1 : 

\(d)\) \(\frac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}.\frac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}=2x\)

\(\Leftrightarrow\)\(\frac{4^5.4}{3^5.3}.\frac{6^5.6}{2^5.2}=2x\)

\(\Leftrightarrow\)\(\frac{4^6}{3^6}.\frac{6^6}{2^6}=2x\)

\(\Leftrightarrow\)\(\frac{2^{12}}{3^6}.\frac{2^6.3^6}{2^6}=2x\)

\(\Leftrightarrow\)\(\frac{2^{12}}{3^6}.\frac{3^6}{1}=2x\)

\(\Leftrightarrow\)\(2^{12}=2x\)

\(\Leftrightarrow\)\(x=\frac{2^{12}}{2}\)

\(\Leftrightarrow\)\(x=2^{11}\)

\(\Leftrightarrow\)\(x=2048\)

Vậy \(x=2048\)

Chúc bạn học tốt ~ 

Bài 1 : 

\(a)\) Ta có : 

\(4+\frac{x}{7+y}=\frac{4}{7}\)

\(\Leftrightarrow\)\(\frac{x}{7+y}=\frac{4}{7}-4\)

\(\Leftrightarrow\)\(\frac{x}{7+y}=\frac{-24}{7}\)

\(\Leftrightarrow\)\(\frac{x}{-24}=\frac{7+y}{7}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có : 

\(\frac{x}{-24}=\frac{7+y}{7}=\frac{x+7+y}{-24+7}=\frac{22+7}{-17}=\frac{29}{-17}=\frac{-29}{17}\)

Do đó : 

\(\frac{x}{-24}=\frac{-29}{17}\)\(\Rightarrow\)\(x=\frac{-29}{17}.\left(-24\right)=\frac{696}{17}\)

\(\frac{7+y}{7}=\frac{-29}{17}\)\(\Rightarrow\)\(y=\frac{-29}{17}.7-7=\frac{-322}{17}\)

Vậy \(x=\frac{696}{17}\) và \(y=\frac{-322}{17}\)

Chúc bạn học tốt ~ 

2.

Ta có 1+2+...+n=n.(n+1):2

=>P=\(1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+...+\)\(\frac{1}{16}.\frac{16.17}{2}\)=1+\(\frac{3}{2}+\frac{4}{2}+...+\frac{17}{2}\)=1+\(\frac{1}{2}.\left(3=4+..=17\right)\)

=1+\(\frac{1}{2}.153=1+\frac{153}{2}=\frac{155}{2}\)