Bài 1:

a) \(8\left(x-2\right)-2\left(3x-4\right)=2\)

\(\Rightarrow2\left[4\left(x-2\right)-\left(3x-4\right)\right]=2\)

\(\Rightarrow4\left(x-2\right)-3x+4=0\)

\(\Rightarrow4x-8-3x+4=0\)

\(\Rightarrow x-4=0\)

\(\Rightarrow x=4\)

b) \(10\left(3x-2\right)-3\left(5x+2\right)+5\left(11-4x\right)=25\)

\(\Rightarrow5\left[2\left(3x-2\right)+11-4x\right]-3\left(5x+2\right)=25\)

\(\Rightarrow5\left(6x-4+11-4x\right)-3\left(5x+2\right)=25\)

\(\Rightarrow5\left(2x+7\right)-3\left(5x+2\right)=25\)

\(\Rightarrow10x+35-15x-6=25\)

\(\Rightarrow-5x+29=25\)

\(\Rightarrow-5x=25-29\)

\(\Rightarrow-5x=-4\)

\(\Rightarrow x=\dfrac{4}{5}\)

c) \(2x\left(x+1\right)-x^2\left(x+2\right)+x^3-x+4=0\)

\(\Rightarrow2x^2+2x-x^3-2x^2+x^3-x+4=0\)

\(\Rightarrow x+4=0\)

\(\Rightarrow x=-4\)

d) \(4x\left(3x+2\right)-6x\left(2x+5\right)+21\left(x-1\right)=0\)

\(\Rightarrow12x^2+8x-12x^2-30x+21x-21=0\)

\(\Rightarrow-x-21=0\)

\(\Rightarrow x=-21\)

Bài 2:

a) \(P=\left(4x^2-3y\right)2y-\left(3x^2-4y\right)3y\)

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

\(P=-x^2y+6y^2\)

Thay x = -1 ; y = 2 vào P ta được

\(P=-\left(-1\right)^2.2+6.2^2\)

\(P=-2+24=22\)

b) \(Q=4x^2\left(5x-3y\right)-x^2\left(4x+y\right)\)

\(Q=20x^3-12x^2y-4x^3-x^2y\)

\(Q=16x^3-13x^2y\)

Thay x = -1 ; y = 2 vào Q ta được

\(Q=16\left(-1\right)^3-13\left(-1\right)^2.2\)

\(Q=-16-26\)

\(Q=-42\)

c) \(H=x\left(x^3-y\right)+x^2\left(y-x^2\right)-y\left(x^2-3x\right)\)

\(H=x^4-xy+x^2y-x^4-x^2y+3xy\)

\(H=2xy\)

Thay x = 1/4 ; y = 2012 vào H ta được

\(H=2.\dfrac{1}{4}.2012\)

\(H=1006\)

1.a)\(8\left(x-2\right)-2\left(3x-4\right)=2\)

\(\Leftrightarrow8x-16-6x+8=2\)

\(\Leftrightarrow2x-8=2\Leftrightarrow2x=10\Leftrightarrow x=5\)

b)\(10\left(3x-2\right)-3\left(5x+2\right)+5\left(11-4x\right)=25\)

\(\Leftrightarrow30x-20-15x-6+55-20x=25\)

\(\Leftrightarrow-5x+29=25\Leftrightarrow-5x=-4\Leftrightarrow x=\dfrac{4}{5}=0,8\)

\(c)2x\left(x+1\right)-x^2\left(x+2\right)+x^3-x+4=0\)

\(\Leftrightarrow2x^2+2x-x^3-2x^2+x^3-x+4=0\)

\(\Leftrightarrow x+4=0\Leftrightarrow x=-4\)

\(d)4x\left(3x+2\right)-6x\left(2x+5\right)+21\left(x-1\right)=0\)

\(\Leftrightarrow12x^2+8x-12x^2-30x+21x-21=0\)

\(\Leftrightarrow-x-21=0\Leftrightarrow-x=21\Leftrightarrow x=-21\)

2.

a)\(P=\left(4x^2-3y\right)2y-\left(3x^2-4y\right)3y\)

\(\Leftrightarrow8x^2y-6y^2-9x^2y-12y^2\)

\(\Leftrightarrow x^2y-18y^2\)

tại x=-1 , y=2

ta có:\(x^2y-18y^2=\left(-1\right)^2.2-18.2^2=2-72=-70\)

vậy \(P=\left(4x^2-3y\right)2y-\left(3x^2-4y\right)3y=-70\) tại x=-1,y=2

b)\(Q=4x^2\left(5x-3y\right)-x^2\left(4x+y\right)\)

\(\Leftrightarrow20x^3-12x^2y-4x^3-x^2y\)

\(\Leftrightarrow17x^3-13x^2y\)

tại x=-1,y=2

ta có:\(17x^3-13x^2y=17\left(-1\right)^3-13\left(-1\right)^2.2=-17-26=-43\)

vậy \(Q=4x^2\left(5x-3y\right)-x^2\left(4x+y\right)=-43\)

c)\(H=x\left(x^3-y\right)+x^2\left(y-x^2\right)-y\left(x^2-3x\right)\)

\(\Leftrightarrow x^4-xy+x^2y-x^3-x^2y+3xy\)

\(\Leftrightarrow x^4+2xy-x^3\)

tại x=1/4 và y=2012

ta có:\(x^4+2xy-x^3=\left(\dfrac{1}{4}\right)^4+2.\dfrac{1}{4}.2012-\left(\dfrac{1}{4}\right)^3\approx1006\)

a. 8(x-2)-2(3x-4)=2

8x-16-6x-8 =2

(8x-6x)-(16-8)=2

2x-2 =2

2x =2+2

2x =4

x =\(\dfrac{4}{2}\)=2

Bài 2:

a) xy = -28

\(\Rightarrow\)x, y \(\in\)Ư(-28)

Ta có: Ư(-28) \(\in\){\(\pm\)1; \(\pm\)2; \(\pm\)4; \(\pm\)7; \(\pm\)14; \(\pm\)28}

Lập bảng:

b) (2x - 1)(4x + 2) = -42

Câu này bạn lập bảng như câu a

c) x + y +xy = 9

\(\Leftrightarrow\)x(y + 1) + (y + 1) = 10

\(\Leftrightarrow\)(x + 1)(y + 1) = 10

\(\Leftrightarrow\)x + 1 và y + 1 \(\in\)Ư(10)

Ta có: Ư(10) \(\in\){\(\pm\)1; \(\pm\)2; \(\pm\)5; \(\pm\)10}

Lập bảng:

d) xy + 3x - 7y = 2

\(\Leftrightarrow\)x(y + 3) - 7y - 21 = -19

\(\Leftrightarrow\)x(y + 3) - 7(y + 3) = -19

\(\Leftrightarrow\)(x - 7)(x + 3) = -19

Tự lập bảng

e) xy - 2x - 3y = 5

\(\Leftrightarrow\)x(y - 2) - 3y + 6 = 11

\(\Leftrightarrow\)x( y - 2) - 3(y - 2) = 11

\(\Leftrightarrow\)(x - 3)(y - 2) = 11

Tự lập bảng

g) xy + 3x -2y = 11

\(\Leftrightarrow\)x(y + 3) - 2y - 6 = 5

\(\Leftrightarrow\)x(y + 3) - 2(y + 3) = 5

\(\Leftrightarrow\)(x - 2)(y + 3) = 5

Tự lập bảng

Bài 1 : Tìm x :

a) (x - 2) (7 - x) > 0

th1 : 

x - 2 > 0 và 7 - x > 0

=> x > 2 và  -x > -7

=> x > 2 và x < 7

=> 2 < x < 7

th2 : 

x - 2 < 0 và 7 - x < 0

=> x < 2 và -x < -7

=> x < 2 và x > 7

=> vô lí

b) (x + 3) (x - 2) < 0 

tương tự câu a

1.

a) như uyên

b) (x+3)(x-2)<0

\(\Leftrightarrow\orbr{\begin{cases}x+3< 0\\x-2< 0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x< -3\\x< 2\end{cases}}\)

Vậy....

2.

a) xy=-28

<=> x;y thuộc Ư(-28)={-1,-2,-4,-7,-14,-28,1,2,4,7,14,28}

Ta có bảng:

Vậy .....

b) (2x-1)(4y+2)=-42

<=> 2x-1 ; 4y+2 thuộc Ư(42)={-1,-2,-3...... bạn tự xét xog lập bảng như trên nhé

1)a)3(2x-1)(3x-1)-(2x-3)(9x-1)=0

<=>18x²-15x+1-18x²+29x-3=0

<=>14x-2=0

<=>14x=2

<=>x=1/7

b)4(x+1)²+(2x-1)²-8(x-1)(x+1)=11

<=>4x²+8x+4+4x²-4x+1-8x²+8=11

<=>4x+13=11

<=>4x=11-13

<=>4x=-2

<=>x=-1/2

c)Sai đề phải là dấu - chứ không phải +

(x-3)(x²+3x+9)-x(x-2)(x+2)=1

<=>x³-27-x³+4x=1

<=>4x=1+27

<=>4x=28

<=>x=7

2)a)(2x-3y)(2x+3y)-4(x-y)²-8xy

=4x²-9y²-4x²+8xy-4y²-8xy

=-13y²

b)(x-2)³-x(x+1)(x-1)+6x(x-3)

=x³-6x²+12x+8-x³+x+6x²-18x

=8-5x

c)(x-2)(x²-2x+4)(x+2)(x²+2x+4)

=(x-2)(x²+2x+4)(x+2)(x²-2x+4)

=(x³-8)(x³+8)

=x⁶-64

Nguyễn Diệu Thảo sap c hk **** cho  Moon Light

câu 1 a. sai .kết quả =0

a.

\(\left\{{}\begin{matrix}\left(x-1\right)^2-\left(y+1\right)^2=0\\x+3y-5=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1-y-1\right)\left(x-1+y+1\right)=0\\x+3y-5=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y-2\right)\left(x+y\right)=0\\x+3y-5=0\end{matrix}\right.\)

TH1: \(\left\{{}\begin{matrix}x-y-2=0\\x+3y-5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{11}{4}\\y=\dfrac{3}{4}\end{matrix}\right.\)

TH2: \(\left\{{}\begin{matrix}x+y=0\\x+3y-5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{2}\\y=\dfrac{5}{2}\end{matrix}\right.\)

b.

\(\left\{{}\begin{matrix}xy-2x-y+2=0\\3x+y=8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\left(y-2\right)-\left(y-2\right)=0\\3x+y=8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)\left(y-2\right)=0\\3x+y=8\end{matrix}\right.\)

TH1:

\(\left\{{}\begin{matrix}x-1=0\\3x+y=8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=5\end{matrix}\right.\)

TH2:

\(\left\{{}\begin{matrix}y-2=0\\3x+y=8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)

c.

\(\left\{{}\begin{matrix}\left(x+y\right)^2-4\left(x+y\right)-12=0\\\left(x-y\right)^2-2\left(x-y\right)=3\end{matrix}\right.\)

Xét pt:

\(\left(x+y\right)^2-4\left(x+y\right)-12=0\)

\(\Leftrightarrow\left(x+y+2\right)\left(x+y-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+y+2=0\\x+y-6=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}y=-x-2\\y=6-x\end{matrix}\right.\)

TH1: \(y=-x-2\) thế vào \(\left(x-y\right)^2-2\left(x-y\right)=3\)

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

\(\Leftrightarrow4x^2+4x-3=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\Rightarrow y=-\dfrac{5}{2}\\x=-\dfrac{3}{2}\Rightarrow y=-\dfrac{1}{2}\end{matrix}\right.\)

TH2: \(y=6-x\) thế vào...

\(\left(2x-6\right)^2-2\left(2x-6\right)=3\)

\(\Leftrightarrow4x^2-28x+45=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\Rightarrow y=\dfrac{7}{2}\\y=\dfrac{9}{2}\Rightarrow y=\dfrac{3}{2}\end{matrix}\right.\)