a) Thay x=2 vào ta có:

\(2^2-4m\cdot2+1=0\\ \Leftrightarrow4-8m+1=0\\ \Leftrightarrow5-8m=0\\ \Leftrightarrow8m=5\\ \Leftrightarrow m=\dfrac{5}{8}\)

b) Thay x=2 vào ta có:

\(3\cdot2^2-5m\cdot2+7\\ \Leftrightarrow12-10m+7=0\\ \Leftrightarrow19-10m=0\\ \Leftrightarrow10m=19\\\Leftrightarrow m=\dfrac{19}{10}\)

a:

Đặt \(x^2-4mx+1=0\left(1\right)\)

Thay x=2 vào (1), ta được:

\(2^2-4m\cdot2+1=0\)

=>\(4-8m+1=0\)

=>5-8m=0

=>8m=5

=>\(m=\dfrac{5}{8}\)

b: Đặt \(3x^2-5mx+7=0\left(2\right)\)

Thay x=2 vào (2), ta được:

\(3\cdot2^2-5m\cdot2+7=0\)

=>12-10m+7=0

=>19-10m=0

=>10m=19

=>\(m=\dfrac{19}{10}\)

a)5x+17-(2x+5)=0

=>5x+17-2x-5=0

=>3x+12=0

=>3x=-12

=>x=-12:3=-4

b)3(1-x)-(5-2x)=0

=>3-3x-5+2x=0

=>-2-x=0

=>x=-2

c)2(x-1)-3(x-2)=0

=>2x-2-3x+6=0

=>-x+4=0

=>x=4

d)(x-3)(2x-5)+(2x-4)(5-2x)=0

=>(x-3)(2x-5)-(2x-4)(2x-5)=0

=>(2x-5)(x-3-2x+4)=0

=>(2x-5)(1-x)=0

TH1: 2x - 5=0=>2x=5=>x=5/2

TH2: 1-x=0=>x=1

a: Đặt 5x+17-(2x+5)=0

=>\(5x+17-2x-5=0\)

=>\(3x+12=0\)

=>\(3x=-12\)

=>\(x=-\dfrac{12}{3}=-4\)

b: Đặt \(3\left(1-x\right)-\left(5-2x\right)=0\)

=>\(3-3x-5+2x=0\)

=>\(-x-2=0\)

=>x+2=0

=>x=-2

c: Đặt \(2\left(x-1\right)-3\left(x-2\right)=0\)

=>\(2x-2-3x+6=0\)

=>4-x=0

=>x=4

d: Sửa đề: (x-3)(2x-5)+(2x-4)*(5-x) 

Đặt \(\left(x-3\right)\left(2x-5\right)+\left(2x-4\right)\left(5-x\right)=0\)

=>\(2x^2-5x-6x+15+10x-2x^2-20+4x=0\)

=>3x-5=0

=>3x=5

=>\(x=\dfrac{5}{3}\)

Đặt 5x+17-(2x+5)=0

=>5x+17-2x-5=0

=>3x+12=0

=>3x=-12

=>\(x=-\dfrac{12}{3}=-4\)

1: \(\left(2x-1\right)^2-\left(4x^2-1\right)=0\)

=>\(\left(2x-1\right)^2-\left(2x-1\right)\left(2x+1\right)=0\)

=>\(\left(2x-1\right)\left(2x-1-2x-1\right)=0\)

=>-2(2x-1)=0

=>2x-1=0

=>\(x=\dfrac{1}{2}\)

2: \(\left(x+2\right)^2-x\left(x-3\right)=2\)

=>\(x^2+4x+4-x^2+3x=2\)

=>7x+4=2

=>7x=-2

=>\(x=-\dfrac{2}{7}\)

3: \(\left(x-5\right)^2-x\left(x+2\right)=5\)

=>\(x^2-10x+25-x^2-2x=5\)

=>-12x+25=5

=>-12x=5-25=-20

=>\(x=\dfrac{20}{12}=\dfrac{5}{3}\)

4: \(\left(x-1\right)^2+x\left(4-x\right)=11\)

=>\(x^2-2x+1+4x-x^2=11\)

=>2x+1=11

=>2x=10

=>x=5

5: \(\left(x-3\right)\left(x+3\right)=\left(x-5\right)^2\)

=>\(x^2-9=x^2-10x+25\)

=>-10x+25=-9

=>-10x=-25-9=-34

=>\(x=\dfrac{34}{10}=\dfrac{17}{5}\)

6: \(\left(2x+1\right)^2-4x\left(x-1\right)=17\)

=>\(4x^2+4x+1-4x^2+4x=17\)

=>8x+1=17

=>8x=16

=>x=2

7: \(\left(3x+1\right)^2-9x\left(x-2\right)=25\)

=>\(9x^2+6x+1-9x^2+18x=25\)

=>24x+1=25

=>24x=24

=>x=1

8: \(\left(3x-2\right)\left(3x+2\right)-9x\left(x-1\right)=0\)

=>\(9x^2-4-9x^2+9x=0\)

=>9x-4=0

=>9x=4

=>\(x=\dfrac{4}{9}\)

9: \(\left(x+2\right)^2-\left(x-2\right)\left(x+2\right)=0\)

=>(x+2)(x+2-x+2)=0

=>4(x+2)=0

=>x+2=0

=>x=-2

10: \(\left(x+2\right)^2-\left(x-3\right)\left(x+3\right)=-3\)

=>\(x^2+4x+4-\left(x^2-9\right)+3=0\)

=>\(x^2+4x+7-x^2+9=0\)

=>4x+16=0

=>4x=-16

=>x=-4

11: \(\left(3x+2\right)^2-\left(3x-5\right)\left(3x+2\right)=0\)

=>(3x+2)(3x+2-3x+5)=0

=>7(3x+2)=0

=>3x+2=0

=>3x=-2

=>\(x=-\dfrac{2}{3}\)

12: \(\left(x+3\right)^2-\left(x+2\right)\left(x-2\right)=4x+17\)

=>\(x^2+6x+9-x^2+4=4x+17\)

=>6x+13=4x+17

=>2x=4

=>x=2

13: \(3\left(x-1\right)^2+\left(x+5\right)\left(-3x+2\right)=-25\)

=>\(3\left(x^2-2x+1\right)+2x-3x^2+10-15x=-25\)

=>\(3x^2-6x+3-3x^2-13x+10=-25\)

=>-19x+13=-25

=>-19x=-38

=>x=2

14: \(\left(x+3\right)^2+\left(x-2\right)^2=2x^2\)

=>\(x^2+6x+9+x^2-4x+4=2x^2\)

=>2x=-13

=>\(x=-\dfrac{13}{2}\)

a) 

\(P=4x^4+y^4\\ =4x^4+4x^2y^2+y^4-4x^2y^2\\ =\left(4x^4+4x^2y^2+y^2\right)-4x^2y^2\\ =\left(2x^2+y^2\right)^2-\left(2xy\right)^2\\ =\left(2x^2-2xy+y^2\right)\left(2x^2+2xy+y^2\right)\) 

b) 

\(Q=x^4+64\\ =x^4+16x^2+64-16x^2\\ =\left(x^4+16x^2+64\right)-16x^2\\ =\left(x^2+8\right)^2-\left(4x\right)^2\\ =\left(x^2-4x+8\right)\left(x^2+4x+8\right)\)

$(-25).(-17).4+(-20)$

$=25.17.4-20$

$=(25.4).17-20$

$=100.17-20$

$=1700-20=1680$

(-25).(-17).4+(-20)

=25.17.4+(-20)

=25.4.17+(-20)

=100.17+(-20)

=1700+(-20)

=1700-20

= 1680

Sửa đề: \(a^2+2ab+b^2-2a-2b+1\)

\(=\left(a^2+2ab+b^2\right)-2\left(a+b\right)+1\)

\(=\left(a+b\right)^2-2\left(a+b\right)\cdot1+1^2\)

\(=\left(a+b-1\right)^2\)

Xét tứ giác ABCD có \(\widehat{ABC}+\widehat{ADC}=180^0\)

nên ABCD là tứ giác nội tiếp

=>\(\widehat{DAC}=\widehat{DBC};\widehat{BAC}=\widehat{BDC}\)

mà \(\widehat{CDB}=\widehat{CBD}\)(CB=CD)

nên \(\widehat{DAC}=\widehat{BAC}\)

=>AC là phân giác của góc BAD

\(x^2+5y^2+2y-4xy-3=0\\ \Rightarrow\left(x^2-4xy+4y^2\right)+\left(y^2+2y+1\right)-4=0\\ \Rightarrow\left(x-2y\right)^2+\left(y+1\right)^2=4\)

Mình nghĩ bạn thiếu đề nhé

Bổ sung đề: Tìm cặp x, y nguyên thỏa mãn

Với x, y nguyên hiển nhiên x-2y và y+1 nguyên

Mà: \(4=0^2+2^2=0^2+\left(-2\right)^2\)

Các trường hợp xảy ra:

TH1: y+1=0 và x-2y=2

=> y=-1 và x=0

TH2: y+1=0 và x-2y=-2

=> y=-1 và x=-4

TH3: y+1=2 và x-2y=0

=> y=1 và x=2

TH4: y+1=-2 và x-2y=0

=> y=-3 và x=-6

Vậy (x;y)=(0;-1);(-4;-1);(2;1);(-6;-3)

Sửa đề: Chứng minh \(\left(a+b\right)^3=a^3+3a^2b+3ab^2+b^3\)

Bài làm:

\(VT=\left(a+b\right)^3=\left(a+b\right)\left(a+b\right)\left(a+b\right)\\ =\left(a^2+ab+ab+b^2\right)\left(a+b\right)\\ =\left(a^2+2ab+b^2\right)\left(a+b\right)\\ =a^3+2a^2b+ab^2+a^2b+2ab^2+b^3\\ =a^3+3a^2b+3ab^2+b^3=VP\left(DPCM\right)\)