Tích 1 số chẵn các thừa số ngâm có dấu ''+''.

nha

 NÓ MANG DẤU  ''  +  ''NHÉ BẠN 

\(xy-2y+x=9\)

\(y\left(x-2\right)+\left(x-2\right)=9-2\)

\(\left(x-2\right)\left(y+1\right)=7\)

tự làm nốt nhé~

Lamtiep bai ban kudo shinichi nk!

Ta co bang:

Vay cac cap (\(\chi\);y) la:(3;8),(1;-8)

HTDT

21-(25-X+23)=  -(35-7)

21-25+X-23=   -28

(-4)+X-23 =  -28

(-4)+X=-28+23 

(-4)+X=-5

X=(-5)-(-4)

X=  -1

a,21-(25-X+23)=-(35-7)\(\Rightarrow21-\left(48-\chi\right)=-28\)

\(\Rightarrow48-\chi=21+28\)

\(\Rightarrow48-\chi=49\Rightarrow\chi=-1\)

HTDT

a) ( -32 ) . ( -56 ) + 32.44

= 32 . 56 + 32 . 44 

= 32  . ( 56 + 44 )

=  32 . 100

= 3200

b) Làm tương tự như phần a nhé !

Kết bạn nhé !

a) \(\left(-32\right)\cdot\left(-56\right)+32\cdot44\)

\(=32\cdot56+32\cdot44\)

\(=32\cdot\left(56+44\right)\)

\(=32\cdot100\)

\(=3200\)

b) \(\left(-59\right)\cdot\left(-56\right)-59\cdot53\)

\(=59\cdot56-59\cdot53\)

\(=59\cdot\left(56-53\right)\)

\(=59\cdot3\)

\(=177\)

chứng minh (X+8) chia hết cho (x+1) hay là tìm x

X+8 CHIA HẾT CHO X+1 

X+1+7 CHIA HẾT CHO X+1

MÀ X+1 CHIA HẾT CHO X+1

=>   7 CHIA HẾT CHO X+1 

=>  X+1 THUỘC Ư (7)

=>X+1 THUỘC {1,7}

=> X THUỘC {0,6}

(47.27).(-13)+13.(-53-27)

=1269.(-13)+13.(-80)

=1269.13.(-1)+13.(-80)

=(1269-80).13.(-1)

=1189.13.(-1)

=15487.(-1)

=-15487

\(\left(47\cdot27\right)\cdot\left(-13\right)+13\cdot\left(-53\cdot27\right)\)

\(=\left(-47\cdot27\right)\cdot13+13\cdot\left(-53\cdot27\right)\)

\(=13\left[\left(-47\cdot27\right)+\left(-53\cdot27\right)\right]\)

\(=13\left\{27\cdot\left[\left(-47\right)+\left(-53\right)\right]\right\}\)

\(=13\cdot\left\{27\cdot\left(-100\right)\right\}\)

\(=13\cdot\left(-2700\right)\)

\(=-35100\)

\(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{100^2}\)

\(=\frac{1}{2^2}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\)

Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)

Ta có: \(\frac{1}{2^2}< \frac{1}{1\cdot2};\frac{1}{3^2}< \frac{1}{2\cdot3};....;\frac{1}{50^2}< \frac{1}{49\cdot50}\)

\(\Rightarrow A< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+....+\frac{1}{49\cdot50}\)

\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)

\(\Rightarrow A< 1-\frac{1}{50}\)

\(\Rightarrow A< 1\Rightarrow1+A< 1+1=2\)

\(\Rightarrow\frac{1}{2^2}\cdot\left(1+A\right)< \frac{1}{2^2}\cdot2=\frac{1}{2}\)(đpcm)

\(A=1-2+3-4+...+97-98+99\)

\(A=\left(1-2\right)+\left(3-4\right)+...+\left(97-98\right)+99\)

\(A=-1+\left(-1\right)+...+\left(-1\right)+99\)

\(A=\left(-1\right).49+99\)

\(A=-49+99\)

\(A=50\)

A = 1 - 2 + 3 - 4 +…+ 97 - 98 + 99

A = ( 1 - 2 ) + ( 3 - 4 ) + ... + ( 97 - 98 ) + 99

A = ( -1 ) + ( -1 ) + ... + ( -1 ) + 99

A = ( -1 ) . 49 + 99

A = 50  

\(\left(-47\right).\left(57-24\right)-57\left(24-47\right)\)

\(=-47.57+47.24-57.24+57.47\)

\(=24.\left(47-57\right)=24.\left(-10\right)=-240\)