1) \(1001^2=\left(1000+1\right)^2=1000^2-1^2=1000000-1=999999\)

2) \(29,9.30,1=\left(30-0,1\right).\left(30+0,1\right)=30^2-0,1^2=900-0,01=899,99\)

Đề bài là gì vậy bạn? Nếu là rút gọn thì mình làm thử nha!

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

\(\Leftrightarrow x^2+2xy+y^2-y^2=x^2+2xy\)

\(\Leftrightarrow x^2-x^2+2xy-2xy+y^2-y^2=0\)

VT: (x+y)²-y²=(x+y-y)(x+y+y)=x(x+2y)

=> VT=VT (đpcm)

Answer:

Có:

\(\frac{x-10}{30}+\frac{x-14}{43}+\frac{x-5}{95}+\frac{x-148}{8}=0\)

\(\Rightarrow\frac{x-10}{30}+\frac{x-14}{43}+\frac{x-5}{95}+\frac{x+100}{8}-6=0\)

\(\Rightarrow\left(\frac{x-10}{30}-3\right)+\left(\frac{x-14}{43}-2\right)+\left(\frac{x-5}{95}-1\right)+\frac{x-100}{8}=0\)

\(\Rightarrow\frac{x-100}{30}+\frac{x-100}{43}+\frac{x-100}{95}+\frac{x-100}{8}=0\)

\(\Rightarrow\left(x-100\right)\left(\frac{1}{30}+\frac{1}{43}+\frac{1}{95}+\frac{1}{9}\right)=0\)

Mà \(\frac{1}{30}+\frac{1}{43}+\frac{1}{95}+\frac{1}{8}\ne0\Leftrightarrow\left(x-100\right)\left(\frac{1}{30}+\frac{1}{43}+\frac{1}{95}+\frac{1}{8}\right)=0\)

\(\Leftrightarrow x-100=0\Leftrightarrow x=100\)

có nhưng có chỗ sai

AB+BC+CD+AD=70

AB+AD+BD=50

CD+BD+BC=60

=>AB+BC+CD+AD+2BD=50+60=110

=>70+2BD=110

=>2BD=40=>BD=20(cm)

a/ Để biểu thức có nghĩa thì: \(\hept{\begin{cases}2x-2\ne0\\2-2x^2\ne0\end{cases}\Rightarrow\hept{\begin{cases}x\ne1\\x\ne-1\end{cases}}}\)

b/ \(C=\frac{x}{2\left(x-1\right)}+\frac{x^2+1}{2\left(1-x^2\right)}=\frac{x}{2\left(x-1\right)}-\frac{x^2+1}{2\left(x+1\right)\left(x-1\right)}\)

       \(=\frac{x\left(x+1\right)-\left(x^2+1\right)}{2\left(x-1\right)\left(x+1\right)}=\frac{x^2+x-x^2-1}{2\left(x-1\right)\left(x+1\right)}=\frac{x-1}{2\left(x-1\right)\left(x+1\right)}\)

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

c/ Có: \(C=-\frac{1}{2}\Leftrightarrow\frac{1}{2\left(x+1\right)}=-\frac{1}{2}\Rightarrow\frac{1}{x+1}=-1\)

               \(\Rightarrow-x-1=1\Rightarrow-x=2\Rightarrow x=-2\)

          Vậy x = -2

B^ + C^ = 180o (bù nhau)

2C^ + C^ = 180o

3C^ = 180o

C^ = 60o

B^ = 2C^ = 2 * 60o = 120o

A^ - D^ = 40o      =>            A^ = 40o + D^ 

A^ + D^ = 180o (bù nhau)

40o + D^ +D^ = 180o

40o + 2D^ = 180o

2D^ = 140o

D^ = 80o

A^ = 40o + D^ = 40o + 80o = 120o

Vậy A^ = 120o

      B^ = 120o

      C^ = 60o

      D^= 80o