Giải chi tiết cho mình nha

\(\dfrac{2}{5}+\dfrac{11}{15}=\dfrac{6}{15}+\dfrac{11}{15}=\dfrac{17}{15}\)

\(\dfrac{7}{8}-\dfrac{7}{9}=\dfrac{63}{72}-\dfrac{56}{72}=\dfrac{7}{72}\)

\(\dfrac{11}{13}\times\dfrac{26}{31}=\dfrac{22}{31}\)

\(\dfrac{16}{24}:4=\dfrac{16}{24}\times\dfrac{1}{4}=\dfrac{4}{24}=\dfrac{1}{6}\)

\(30:\dfrac{6}{5}=30\times\dfrac{5}{6}=25\)

\(\dfrac{9}{16}:4=\dfrac{9}{16}\times\dfrac{1}{4}=\dfrac{9}{64}\)

\(\dfrac{1}{2}:\dfrac{1}{3}\times\dfrac{8}{15}=\dfrac{1}{2}\times3\times\dfrac{8}{15}=\dfrac{4}{5}\)

2/5 + 11/15 = 17/15
7/8 - 7/9 = 7/72
11/13 x 26/31 = 22/31
16/24 : 4 = 1/6
30 : 6/5 = 25
9/16 : 4 = 9/64
1/2 : 1/3 x 8/15 = 3/2 x 8/15 = 4/5

h) \(=\frac{x+1}{32}\)

câu hỏi tương tự

1) 

=a^4+2a^2+1-a^2

=(a^2+1)^2-a^2

=(a^2-a+1)(a^2+a+1)

2)

=a^4+4b^4-4a^2b^2

=(a^2+2b^2)^2-4a^2b^2

=(a^2-2ab+2b^2)(a^2+2ab+2b^2)

3)

=(8x^2+1)^2-16x^2

=(8x^2-4x+1)(8x^2+4x+1).

4)

=x^5+x^4+x^3-x^3+1

=x^2(x^2+x+1)-(x-1)(x^2+x+1)

=(x^2-x+1)(x^2+x+1)

5).

=x^7-x+x^2+x+1

=x(x^6-1)+x^2+x+1

=x(x^3-1)(x^3+1)+x^2+x+1

=x(x-1)(x^2+x+1)(x^3+1)+x^2+x+1

=(x^2+x+1)[(x^2-x)(x^3+1)+1]

6)

=x^8-x^2+x^2+x+1

=x^2(x-1)(x^2+x+1)(x^3+1)+x^2+x+1

Xong nhóm x^2+x+1 vào.

7)

=x^4-(2x-1)^2

=(x^2-2x+1)(x^2+2x-1)

8)

=(a^8+b^8)^2-a^8b^8

=(a^8-a^4b^4+b^8)(a^8+a^4b^4+b^8).

Bài 1 :

a ) Ta có :

\(\left(x+y\right)^2=x^2+y^2+2xy=20+16=36\)

b ) Ta có :

\(x^2+y^2=\left(x+y\right)^2-2xy=64-30=34\)

1: =(x+y-3x)(x+y+3x)

=(-2x+y)(4x+y)

2: =(3x-1-4)(3x-1+4)

=(3x+3)(3x-5)

=3(x+1)(3x-5)

3: =(2x)^2-(x^2+1)^2

=-[(x^2+1)^2-(2x)^2]

=-(x^2+1-2x)(x^2+1+2x)

=-(x-1)^2(x+1)^2

4: =(2x+1+x-1)(2x+1-x+1)

=3x(x+2)

5: =[(x+1)^2-(x-1)^2][(x+1)^2+(x-1)^2]

=(2x^2+2)*4x

=8x(x^2+1)

6: =(5x-5y)^2-(4x+4y)^2

=(5x-5y-4x-4y)(5x-5y+4x+4y)

=(x-9y)(9x-y)

7: =(x^2+xy+y^2+xy)(x^2+xy-y^2-xy)

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

=(x+y)^3*(x-y)

8: =(x^2+4y^2-20-4xy+16)(x^2+4y^2-20+4xy-16)

=[(x-2y)^2-4][(x+2y)^2-36]

=(x-2y-2)(x-2y+2)(x+2y-6)(x+2y+6)

1, a⁴ + a² + 1 

= a⁴ + 2a² + 1 - a² 

= (a²)² + 2a² + 1 - a² 

= (a² + 1)² - a² 

= (a² + 1 - a)(a² + 1 + a)

2, a⁴ + 4b⁴ 

= (a²)² + 2. a² . b² + (2b)² - a² . b² 

= (a² + 2b)² - (ab)² 

= (a² + 2b - ab)(a² + 2b + ab)

3, 64x⁴ + 1 

= (8x²)²​ + 16x²​ + 1 - 16x²​ 

= (8x² + 1)²​ - (4x)²​ 

= (8x² + 1 - 4x)(8x² + 1 + 4x)

4, x⁵ + x⁴ + 1 

= x⁵ + x⁴ + x³ - x³ - x² - x + x + x² + 1 

= (x⁵ + x⁴ + x³) - (x³ + x² + x) + (x + x² + 1)

= x³(x² + x + 1) - x(x² + x + 1) + (x² + x + 1)

= (x² + x + 1)(x³ - x + 1)

5, x⁷ + x² + 1 

= x⁷ – x + x² + x + 1

= x(x⁶ – 1) + (x² + x + 1) 

= x(x³ – 1)(x³ + 1) + (x² + x + 1)

= x(x³ + 1)(x – 1) (x² + x + 1) + (x² + x + 1) 

= (x² + x + 1)[ x(x³ + 1)(x – 1) + 1]

= (x² + x + 1)(x⁵ – x⁴ + x³ – x² + x – 1)

6, x⁸ + x + 1 

= x⁸ + x⁷ + x⁶ - x⁷ - x⁶ - x⁵ + x⁵ + x⁴ + x³ - x⁴ - x³ - x² + x² + x + 1

= (x⁸ + x⁷ + x⁶) -  (x⁷ + x⁶ + x⁵) + (x⁵ + x⁴ + x³ ) - (x⁴ + x³ + x²) + (x² + x + 1)

= x⁶(x² + x + 1) - x⁵(x² + x + 1) + x³(x² + x + 1) - x²(x² + x + 1) + (x² + x + 1)

= (x² + x + 1)(x⁶ - x⁵ + x³ - x² + 1)

7, x⁴ - 4x² + 4x - 1 

= x⁴ - (4x² - 4x + 1)

= (x²)² - (2x - 1)²

= (x² - 2x + 1)(x² + 2x - 1)

= (x - 1)² (x² + 2x - 1)

8, a¹⁶ + a⁸b⁸ + b¹⁶

=  (a¹⁶ + 2a⁸b⁸ + b¹⁶) - a⁸b⁸ 

= (a⁸ + b⁸)² - (a⁴b⁴)²

= (a⁸ + b⁸ - a⁴b⁴)(a⁸ + b⁸ + a⁴b⁴)

= (a⁸ + b⁸ - a⁴b⁴)[(a⁸ + b⁸ + 2a⁴b⁴) - a⁴b⁴]

= (a⁸ + b⁸ - a⁴b⁴)[(a⁴ + b⁴)² - (a²b²)²]

= (a⁸ + b⁸ - a⁴b⁴)(a⁴ + b⁴ - a²b²)(a⁴ + b⁴ + a²b²)

= (a⁸ + b⁸ - a⁴b⁴)(a⁴ + b⁴ - a²b²)[(a⁴ + b⁴ + 2a²b²) - a²b²]

= (a⁸ + b⁸ - a⁴b⁴)(a⁴ + b⁴ - a²b²)[(a² + b²) - (ab)²]

= (a⁸ + b⁸ - a⁴b⁴)(a⁴ + b⁴ - a²b²)(a² + b² - ab)(a² + b² + ab)

a)    \(\frac{3}{16}+\frac{4}{15}+\frac{5}{16}+\frac{1}{15}\)

\(=\left(\frac{3}{16}+\frac{5}{16}\right)+\left(\frac{4}{15}+\frac{1}{15}\right)\)

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

\(=\frac{5}{6}\)

b)   \(\frac{6}{7}\times\frac{8}{15}\times\frac{7}{6}\times\frac{15}{16}\)

\(=\left(\frac{6}{7}\times\frac{7}{6}\right)\times\left(\frac{8}{15}\times\frac{15}{16}\right)\)

\(=1\times\frac{1}{2}=\frac{1}{2}\)

c)   \(\frac{19}{20}\times\frac{13}{21}+\frac{9}{20}\times\frac{8}{21}\)

\(=\frac{19\times13}{20\times21}+\frac{9\times8}{20\times21}\)

\(=\frac{247}{420}+\frac{72}{420}\)

\(=\frac{319}{420}\)

còn phần D đâu bn 

tích cho chị nha

 câu a)     = 5/6

 câu b = 1/2

 câu c = 319/420

\(\dfrac{3}{2}\times\dfrac{9}{4}\times\dfrac{81}{16}=\dfrac{3}{2}\times\left(\dfrac{3}{2}\right)^2\times\left(\dfrac{3}{2}\right)^4=\left(\dfrac{3}{2}\right)^7\)

\(\left(\dfrac{1}{2}\right)^7\times8\times32\times2^8=\left(\dfrac{1}{2}\right)^7\times2^3\times2^5\times2^8=\left(\dfrac{1}{7}\right)^7\times2^{16}\)

\(\left(-\dfrac{1}{7}\right)^4\times125\times5=\left(-\dfrac{1}{7}\right)^4\times5^3\times5=\left(-\dfrac{1}{7}\right)^4\times5^4\)

\(4\times32:\left(2^3\times\dfrac{1}{16}\right)=2^2\times2^5:2^3:2^{-4}=2^0\)

\(\left(\dfrac{1}{7}\right)^2\times\dfrac{1}{7}\times49=\left(\dfrac{1}{7}\right)^3\times7^3=1^3\)
 

6, \(\dfrac{3}{2}\times\dfrac{9}{4}\times\dfrac{81}{16}=\dfrac{3}{2}\times\left(\dfrac{3}{2}\right)^2\times\left(\dfrac{3}{2}\right)^4\)

7,\(\left(\dfrac{1}{2}\right)^7\times8\times32\times2^8=\left(\dfrac{1}{2}\right)^7\times2^3\times2^5\times2^8\)

8,\(\left(-\dfrac{1}{7}\right) ^4\times125\times5=\left(\dfrac{1}{7}\right)^4\times5^3\times5\)

9,\(4\times32:\left(2^3\times\dfrac{1}{16}\right)=2^2\times2^5:\left[2^3\times\left(\dfrac{1}{2}\right)^4\right]\)

10, \(\left(\dfrac{1}{7}\right)^2\times\dfrac{1}{7}\times49=\left(\dfrac{1}{7}\right)^2\times\dfrac{1}{7}\times7^2\)

6) \(...=\dfrac{3}{2}.\left(\dfrac{3}{2}\right)^2.\left(\dfrac{3}{2}\right)^4=\left(\dfrac{3}{2}\right)^7\)

7) \(...=2^{-7}.2^3.2^5.2^8=2^9\)

8) \(...=\left(\dfrac{1}{7}\right)^4.5^3.5=\left(\dfrac{1}{7}\right)^4.5^4\)

9) \(...=2^2.2^5:\left(2^3.2^{-4}\right)=2^8\)

10) \(...=\left(\dfrac{1}{7}\right)^3.\left(\dfrac{1}{7}\right)^2=\left(\dfrac{1}{7}\right)^5\)

1) ta có \(\left(x+y\right)^2=x^2+2xy+y^2.\)

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

                                \(=20+2.8\)(theo giả thiết x^2+y^2=20 , xy=8)

                                \(=36\)

Vậy với x^2+y^2=20, xy=8 thì (x+y)^2=36

2) \(M=\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

 \(\Rightarrow3M=3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

  \(\Leftrightarrow3M=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

 \(\Leftrightarrow3M=\left[\left(2^2\right)^2-1^2\right]\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(\Leftrightarrow3M=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

 \(\Leftrightarrow3M=\left[\left(2^4\right)^2-1^2\right]\left(2^8+1\right)\left(2^{16}+1\right)\)

\(\Leftrightarrow3M=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(\Leftrightarrow3M=\left[\left(2^8\right)^2-1^2\right]\left(2^{16}+1\right)\)

\(\Leftrightarrow3M=\left(2^{16}-1\right)\left(2^{16}+1\right)\)

\(\Leftrightarrow3M=\left(2^{16}\right)^2-1^2\)

\(\Leftrightarrow3M=2^{32}-1\)

\(\Rightarrow M=\frac{2^{32}-1}{3}\)

RÚT GỌN BIỂU THỨC N BẠN LÀM TƯƠNG TỰ NHA 

\(N=16\left(7^2+1\right)\left(7^4+1\right)\left(7^8+1\right)\left(7^{16}+1\right)\)

 \(\Rightarrow3N=48\left(7^2+1\right)\left(7^4+1\right)\left(7^8+1\right)\left(7^{16}+1\right)\)

\(\Leftrightarrow3N=\left(7^2-1\right)\left(7^2+1\right)\left(7^4+1\right)\left(7^8+1\right)\left(7^{16}+1\right)\)

\(...\)

\(...\)

Kết quả rút gọn \(N=\frac{7^{32}-1}{3}\)