Bài 8:

1: \(\left(x+y\right)^2-\left(x-y\right)^2\)

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

\(=2x\cdot2y=4xy\)

2: \(\left(2x+3\right)^2-3x\left(2x+1\right)\)

\(=4x^2+12x+9-6x^2-3x\)

\(=-2x^2+9x+9\)

3: \(\left(4-2x\right)\left(4+2x\right)-4x\left(2x+3\right)\)

\(=4^2-\left(2x\right)^2-8x^2-12x\)

\(=16-4x^2-8x^2-12x=-12x^2-12x+16\)

4: \(2\left(x+y\right)\left(x-y\right)+\left(x+y\right)^2-2x^2\)

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

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

5: \(\left(3x+4\right)\left(3x-2\right)-\left(3x+1\right)^2\)

\(=9x^2-6x+12x-8-9x^2-6x-1\)

=-9

6: \(4x\left(x-3\right)-\left(2x-1\right)\left(2x+1\right)\)

\(=4x^2-12x-\left(4x^2-1\right)\)

\(=4x^2-12x-4x^2+1=-12x+1\)

7: \(\dfrac{3}{2}x^2-\left(x-1\right)\left(x+1\right)+3x\)

\(=\dfrac{3}{2}x^2+3x-\left(x^2-1\right)\)

\(=\dfrac{3}{2}x^2+3x-x^2+1=\dfrac{1}{2}x^2+3x+1\)

8: \(2\left(5-x\right)\left(5+x\right)-\left(2x+3\right)^2-x\left(3x+2\right)\)

\(=2\left(25-x^2\right)-4x^2-12x-9-3x^2-2x\)

\(=2\left(25-x^2\right)-7x^2-14x-9\)

\(=50-2x^2-7x^2-14x-9=-9x^2-14x+41\)

Giúp tớ nhanh vs ạ

Bài 8:

\(1)\left(x+y\right)^2-\left(x-y\right)^2\\ =\left(x^2+2xy+y^2\right)-\left(x^2-2xy+y^2\right)\\ =x^2+2xy+y^2-x^2+2xy-y^2\\ =4xy\\ 2)\left(2x+3\right)^2-3x\left(2x+1\right)\\ =\left(4x^2+12x+9\right)-\left(6x^2+3x\right)\\ =4x^2+12x+9-6x^2-3x\\ =-2x^2+9x+9\\ 3)\left(4-2x\right)\left(4+2x\right)-4x\left(2x+3\right)\\ =\left[4^2-\left(2x\right)^2\right]-\left(8x^2+12x\right)\\ =16-4x^2-8x^2-12x\\ =16-12x^2-12x\\ 4)2\left(x+y\right)\left(x-y\right)+\left(x+y\right)^2-2x^2\\ =2\left(x^2-y^2\right)+\left(x^2+2xy+y^2\right)-2x^2\\ =2x^2-2y^2+x^2+2xy+y^2-2x^2\\ =x^2+2xy-y^2\)

Bài 8:

1: \(\left(x+y\right)^2-\left(x-y\right)^2\)

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

\(=2x\cdot2y=4xy\)

2: \(\left(2x+3\right)^2-3x\left(2x+1\right)\)

\(=4x^2+12x+9-6x^2-3x\)

\(=-2x^2+9x+9\)

3: \(\left(4-2x\right)\left(4+2x\right)-4x\left(2x+3\right)\)

\(=4^2-\left(2x\right)^2-8x^2-12x\)

\(=16-4x^2-8x^2-12x=-12x^2-12x+16\)

4: \(2\left(x+y\right)\left(x-y\right)+\left(x+y\right)^2-2x^2\)

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

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

5: \(\left(3x+4\right)\left(3x-2\right)-\left(3x+1\right)^2\)

\(=9x^2-6x+12x-8-9x^2-6x-1\)

=-9

6: \(4x\left(x-3\right)-\left(2x-1\right)\left(2x+1\right)\)

\(=4x^2-12x-\left(4x^2-1\right)\)

\(=4x^2-12x-4x^2+1=-12x+1\)

7: \(\dfrac{3}{2}x^2-\left(x-1\right)\left(x+1\right)+3x\)

\(=\dfrac{3}{2}x^2+3x-\left(x^2-1\right)\)

\(=\dfrac{3}{2}x^2+3x-x^2+1=\dfrac{1}{2}x^2+3x+1\)

8: \(2\left(5-x\right)\left(5+x\right)-\left(2x+3\right)^2-x\left(3x+2\right)\)

\(=2\left(25-x^2\right)-4x^2-12x-9-3x^2-2x\)

\(=2\left(25-x^2\right)-7x^2-14x-9\)

\(=50-2x^2-7x^2-14x-9=-9x^2-14x+41\)

11.

a) 

\(A=\left(x+1\right)\left(x^2-x+1\right)-\left(x-1\right)\left(x^2+x+1\right)\\=\left(x+1\right)\left(x^2-x\cdot1+1^2\right)-\left(x-1\right)\left(x^2+x\cdot1+1^2\right)\\ =\left(x^3+1^3\right)-\left(x^3-1^3\right)\\ =x^3+1-x^3+1\\ =2\)

=> Giá trị của bt không phụ thuộc vào biến 

b) 

\(B=\left(2x+6\right)\left(4x^2-12x+36\right)-8x^3+10\\ =\left(2x+6\right)\left[\left(2x\right)^2-2x\cdot6+6^2\right]-8x^3+10\\ =\left[\left(2x\right)^3+6^3\right]-8x^3+10\\ =\left(8x^3+216\right)-8x^3+10\\ =8x^3+216-8x^3+10\\ =226\)

=> Giá trị của bt không phụ thuộc vào biến

6.

\(a)\left(x+1\right)^3=x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=x^3+3x^2+3x+1\\ b)\left(2x+3\right)^3=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot3+3\cdot2x\cdot3^2+3^3=8x^3+36x^2+54x+27\\ c)\left(x^2+2\right)^3=\left(x^2\right)^3+3\cdot\left(x^2\right)^2\cdot2+3\cdot x^2\cdot2^2+2^3=x^6+6x^4+12x^2+8\\ d)\left(2x+5y\right)^3=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot5y+3\cdot2x\cdot\left(5y\right)^2+\left(5y\right)^3=8x^3+60x^2y+150xy^2+125y^3\\ e.\left(x+\dfrac{1}{2}\right)^3=x^3+3\cdot x^2\cdot\dfrac{1}{2}+3\cdot x\cdot\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3=x^3+\dfrac{3}{2}x^2+\dfrac{3}{4}x+\dfrac{1}{8}\\ g.\left(\dfrac{1}{2}x+y^2\right)=\left(\dfrac{1}{2}x\right)^3+3\cdot\left(\dfrac{1}{2}x\right)^2\cdot y^2+3\cdot\dfrac{1}{2}x\cdot\left(y^2\right)^2+\left(y^2\right)^3\\ =\dfrac{x^3}{8}+\dfrac{3}{4}x^2y^2+\dfrac{3}{2}xy^4+y^6\\ h.\left(x^2-2\right)^3=\left(x^2\right)^3-3\cdot\left(x^2\right)^2\cdot2+3\cdot x^2\cdot2^2-2^3=x^6-6x^4+12x^2-8\)

Bài 14:

1: \(A=x^2-x+3\)

\(=x^2-x+\dfrac{1}{4}+\dfrac{11}{4}\)

\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{11}{4}>=\dfrac{11}{4}\forall x\)

Dấu '=' xảy ra khi x-1/2=0

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

2: \(B=x^2+x+1\)

\(=x^2+x+\dfrac{1}{4}+\dfrac{3}{4}\)

\(=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>=\dfrac{3}{4}\forall x\)

Dấu '=' xảy ra khi \(x+\dfrac{1}{2}=0\)

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

3: \(C=x^2-4x+1\)

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

\(=\left(x-2\right)^2-3>=-3\forall x\)

Dấu '=' xảy ra khi x-2=0

=>x=2

4: \(D=x^2-5x+7\)

\(=x^2-2\cdot x\cdot\dfrac{5}{2}+\dfrac{25}{4}+\dfrac{3}{4}\)

\(=\left(x-\dfrac{5}{2}\right)^2+\dfrac{3}{4}>=\dfrac{3}{4}\forall x\)

Dấu '=' xảy ra khi \(x-\dfrac{5}{2}=0\)

=>\(x=\dfrac{5}{2}\)

5: \(E=x^2+2x+2\)

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

Dấu '=' xảy ra khi x+1=0

=>x=-1

6: \(F=x^2-3x+1\)
\(=x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{5}{4}\)

\(=\left(x-\dfrac{3}{2}\right)^2-\dfrac{5}{4}>=-\dfrac{5}{4}\forall x\)

Dấu '=' xảy ra khi \(x-\dfrac{3}{2}=0\)

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

7: \(G=x^2+3x+3\)

\(=x^2+2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{3}{4}\)

\(=\left(x+\dfrac{3}{2}\right)^2+\dfrac{3}{4}>=\dfrac{3}{4}\forall x\)

Dấu '=' xảy ra khi x+3/2=0

=>x=-3/2

8: \(H=3x^2+3-5x\)

\(=3\left(x^2-\dfrac{5}{3}x+1\right)\)

\(=3\left(x^2-2\cdot x\cdot\dfrac{5}{6}+\dfrac{25}{36}+\dfrac{11}{36}\right)\)

\(=3\left(x-\dfrac{5}{6}\right)^2+\dfrac{11}{12}>=\dfrac{11}{12}\forall x\)

Dấu '=' xảy ra khi x-5/6=0

=>x=5/6

9: \(I=4x+2x^2+3\)

\(=2\left(x^2+2x+\dfrac{3}{2}\right)\)

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

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

Dấu '=' xảy ra khi x+1=0

=>x=-1

10: \(K=4x^2+3x+2\)

\(=\left(2x\right)^2+2\cdot2x\cdot\dfrac{3}{4}+\dfrac{9}{16}+\dfrac{23}{16}\)

\(=\left(2x+\dfrac{3}{4}\right)^2+\dfrac{23}{16}>=\dfrac{23}{16}\forall x\)

Dấu '=' xảy ra khi 2x+3/4=0

=>x=-3/8

11: M=(x-1)(x-3)+11

\(=x^2-4x+3+11=x^2-4x+14\)

\(=x^2-4x+4+10=\left(x-2\right)^2+10>=10\forall x\)

Dấu '=' xảy ra khi x-2=0

=>x=2

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

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

\(=2x^2-10x+13\)

\(=2\left(x^2-5x+\dfrac{13}{2}\right)=2\left(x^2-5x+\dfrac{25}{4}+\dfrac{1}{4}\right)\)

\(=2\left(x-\dfrac{5}{2}\right)^2+\dfrac{1}{2}>=\dfrac{1}{2}\forall x\)

Dấu '=' xảy ra khi x-5/2=0

=>x=5/2

Gọi mẫu số là x

(ĐIều kiện: \(x\ne0\))

Vì phân số nhỏ hơn 1 nên mẫu số>tử số

=>Mẫu số>32/2=16

Tử số là 32-x

Mẫu số khi tăng thêm 10 đơn vị là x+10

Tử số khi giảm đi một nửa là \(\dfrac{32-x}{2}\)

Phân số mới là 2/17 nên \(\dfrac{32-x}{2}:\left(x+10\right)=\dfrac{2}{17}\)

=>\(\dfrac{32-x}{2x+20}=\dfrac{2}{17}\)

=>17(32-x)=2(2x+20)

=>544-17x=4x+40

=>-21x=40-544=-504

=>x=24

Tử số là 32-24=8

Vậy: Phân số cần tìm là \(\dfrac{8}{24}\)

(x-y)(x-2)=11

=>\(\left(x-y;x-2\right)\in\left\{\left(1;11\right);\left(11;1\right);\left(-1;-11\right);\left(-11;-1\right)\right\}\)
=>\(\left(x-2;x-y\right)\in\left\{\left(1;11\right);\left(11;1\right);\left(-1;-11\right);\left(-11;-1\right)\right\}\)

=>\(\left(x;x-y\right)\in\left\{\left(3;11\right);\left(13;1\right);\left(1;-11\right);\left(-9;-1\right)\right\}\)

=>\(\left(x;y\right)\in\left\{\left(3;-8\right);\left(13;12\right);\left(1;12\right);\left(-9;-8\right)\right\}\)

Gọi H là giao điểm của CN với BM

Xét ΔHCB có

CM,BN là các đường cao

CM cắt BN tại A

Do đó: A là trực tâm của ΔHCB

=>HA\(\perp\)CB tại K

Xét ΔBKA vuông tại K và ΔBNC vuông tại N có

\(\widehat{CBN}\) chung

Do đó: ΔBKA~ΔBNC

=>\(\dfrac{BK}{BN}=\dfrac{BA}{BC}\)

=>\(BN\cdot BA=BK\cdot BC\)

Xét ΔCKA vuông tại K và ΔCMB vuông tại M có

\(\widehat{KCA}\) chung

Do đó: ΔCKA~ΔCMB

=>\(\dfrac{CK}{CM}=\dfrac{CA}{CB}\)

=>\(CM\cdot CA=CK\cdot CB\)

\(BA\cdot BN+CA\cdot CM\)

\(=BC\cdot BK+BC\cdot CK=BC\left(BK+CK\right)=BC^2\)

Xét ΔOAB và ΔOCD có

\(\widehat{OAB}=\widehat{OCD}\)(AB//CD)

\(\widehat{AOB}=\widehat{COD}\)(hai góc đối đỉnh)

Do đó: ΔOAB~ΔOCD

=>\(\dfrac{OA}{OC}=\dfrac{OB}{OD}\)(1)

Xét ΔOBP và ΔODQ có

\(\widehat{OBP}=\widehat{ODQ}\)(BP//DQ)

\(\widehat{BOP}=\widehat{DOQ}\)(hai góc đối đỉnh)

Do đó: ΔOBP~ΔODQ

=>\(\dfrac{OB}{OD}=\dfrac{OP}{OQ}\left(2\right)\)

Xét ΔOAM và ΔOCN có

\(\widehat{OAM}=\widehat{OCN}\)(AM//CN)

\(\widehat{AOM}=\widehat{CON}\)(hai góc đối đỉnh)

Do đó: ΔOAM~ΔOCN

=>\(\dfrac{OA}{OC}=\dfrac{OM}{ON}\left(3\right)\)

Từ (1),(2),(3) suy ra \(\dfrac{OP}{OQ}=\dfrac{OM}{ON}\)

=>\(OP\cdot ON=OM\cdot OQ\)