a)Ta có:A2=B4(đối đỉnh)

              B2=B4(đối đỉnh)

Lại có:A1=B1 nên A2=B2

b) Ta có:A3=B1(đối đỉnh)

               B3=B1(đối đỉnh)

Lại có:A2=B2 nên A3=B3

c)Ta có :B3+B2=180° (kề bù)

Lại có :B3=A3

nên A3+B2=180°

a,\(\left[196-4\cdot\left(5+3x\right)\right]+16=96\)

\(\left[196-4\cdot\left(5+3x\right)\right]=96-16\)

\(\left[196-4\cdot\left(5+3x\right)\right]=80\)

\(4\cdot\left(5+3x\right)=196-80\)

\(4\cdot\left(5+3x\right)=116\)

\(5+3x=116\div4\)

\(5+3x=29\)

\(3x=29-5\)

\(3x=24\)

\(x=24\div3\)

\(x=8\)

b,\(\left[6+105\div\left(35-7x\right)\right]\cdot3-7=26\)

\(\left[6+105\div\left(35-7x\right)\right]\cdot3=26+7\\ \left[6+105\div\left(35-7x\right)\right]\cdot3=33\\ \left[6+105\div\left(35-7x\right)\right]=33\div3\\ \left[6+105\div\left(35-7x\right)\right]=11\)

\(\left[105\div\left(35-7x\right)\right]=11-6\\\left[105\div\left(35-7x\right)\right]=5\\ \left(35-7x\right)=105\div5\\ \left(35-7x\right)=21\\ 7x=35-21\\7x=14\\ x=14\div7\\ x=2\)

\(\dfrac{205}{206}+\dfrac{206}{207}+\dfrac{207}{205}\)

\(=1\) ( vì có các số giống nhau )

b,\(\dfrac{45678\cdot1997+3333}{1998+45678-12345}\)

\(=\dfrac{45678\cdot1998+3333}{1998+45678-12345}\)

\(=\dfrac{1111}{4115}\)

đề câu f sao lạ vậy 2345 hay 2545 vậy ???

a.\(\dfrac{1996\cdot2011-1000}{996+2010\cdot1996}\)

\(=\dfrac{1996\cdot2011-1-1000}{996+4+2010\cdot1996}\)

\(=1\)

a)\(1000\div125\cdot35\)

=\(8\cdot35\)

\(=280\)

b)\(234\cdot2+169\div3\)

\(=468+\dfrac{196}{3}\)

\(=\dfrac{1600}{3}\)

c)\(10^2\cdot3^2+5^3\)

\(=900+125\)

\(=1025\)

d)\(12^4\div12^3-7^3\div7\)

\(=12-49\)

\(=\left(-37\right)\)

e)\(\left(21+19\right)^4\div40^26+31-1600\)

\(=40^4\div40^2+31-1600\)

\(=40^2+31-1600\)

\(=1600-1600+31\)

\(=31\)

spam bạn ???

\(A=1+3+3^2+3^3+3^4+...+3^{100}-3^{101}\div2+5\)

\(3A=3+1+3+3^2+3^3+...+3^{100}\div2+5\)

\(3A-A=\left(1+3+3^2+3^3+3^4+...+3^{100}+3^{101}\right)-\left(3+1+3+3^2+3^3+...+3^{100}\right)\div2+5\)

\(2A=1+3+3^2+3^3+3^4+...+3^{100}+3^{101}-3-1-3-3^2-3^3-...-3^{100}\div2+5\)

\(2A=3-3^{101}\div2+5\)

\(2A=3-3^{101}\div2^{101}+5\)

\(2A=3-1+5\)

\(2A=7\)

\(\Rightarrow A=14\)

Ta có: \(x^2-2xy=4\) để bằng 4 thì \(2xy=0\) nên\(x=2;y=0\)

Vậy x=2 và y=0.