19

Từ pt đầu ta có:

\(x^2-xy-2xy+2y^2=0\)

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

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

\(\Rightarrow\left[{}\begin{matrix}x=y\\x=2y\end{matrix}\right.\)

TH1: \(x=y\) thế xuống pt dưới:

\(y^2-y-y^2=1\Rightarrow y=-1\Rightarrow x=-1\)

TH2: \(x=2y\) thế xuống pt dưới:

\(\left(2y\right)^2-2y-y^2=1\Leftrightarrow3y^2-2y-1=0\)

\(\Rightarrow\left[{}\begin{matrix}y=1\Rightarrow x=2\\y=-\dfrac{1}{3}\Rightarrow x=-\dfrac{2}{3}\end{matrix}\right.\)

Vậy nghiệm của hệ là: \(\left(x;y\right)=\left(-1;-1\right);\left(1;2\right);\left(-\dfrac{1}{3};-\dfrac{2}{3}\right)\)

21.

Từ pt đầu:

\(xy+2=2x+y\Leftrightarrow xy-y+2-2x=0\)

\(\Leftrightarrow y\left(x-1\right)-2\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(y-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)

TH1: \(x=1\) thế xuống pt dưới:

\(2y+y^2+3y=6\Leftrightarrow y^2+5y-6=0\)

\(\Rightarrow\left[{}\begin{matrix}y=1\\y=-6\end{matrix}\right.\)

TH2: \(y=2\) thế xuông pt dưới

\(4x+4+6=6\Rightarrow x=-1\)

Vậy nghiệm của pt là: \(\left(x;y\right)=\left(1;1\right);\left(1;-6\right);\left(-1;2\right)\)

Ai giúp em với ạ

Bài 3:

a: \(36x^2-49=0\)

=>\(36x^2=49\)

=>\(x^2=\frac{49}{36}\)

=>\(\left[\begin{array}{l}x=\frac76\\ x=-\frac76\end{array}\right.\)

b: \(\left(x-1\right)\left(x+2\right)=x+2\)

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

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

=>\(\left[\begin{array}{l}x+2=0\\ x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-2\\ x=2\end{array}\right.\)

c: \(x^2\left(x+1\right)+2x\left(x+1\right)=0\)

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

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

=>\(\left[\begin{array}{l}x=0\\ x+1=0\\ x+2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=-1\\ x=-2\end{array}\right.\)

d: x(2x-3)-3(3-2x)=0

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

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

=>\(\left[\begin{array}{l}2x-3=0\\ x+3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=3\\ x=-3\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac32\\ x=-3\end{array}\right.\)

e: \(2x^3\left(2x-3\right)-x^2\left(4x^2-6x+2\right)=0\)

=>\(4x^4-6x^3-4x^4+6x^3-2x^2=0\)

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

=>\(x^2=0\)

=>x=0

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

=>\(x^2-4x+4-x^2-6x-9=5+4x+4\)

=>-10x-5=4x+9

=>-14x=14

=>x=-1

Bài 4:

a: \(5a^2b+10ab^2-5ab=5ab\cdot a+5ba\cdot2b-5ab\cdot1\)

=5ab(a+2b-1)

b: \(4x^2+8xy-3x-6y\)

=4x(x+2y)-3(x+2y)

=(x+2y)(4x-3)

c: 10x(x-y)-6y(y-x)

=10x(x-y)+6y(x-y)

=(x-y)(10x+6y)

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

d: \(3x^2-6xy+3y^2-12\)

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

\(=3\left\lbrack\left(x-y\right)^2-4\right\rbrack\)

=3(x-y-2)(x-y+2)

e: \(16-x^2+2xy-y^2\)

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

\(=16-\left(x-y\right)^2=\left(4-x+y\right)\left(4+x-y\right)\)

f: \(3x^2-5x+2\)

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

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

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

g: \(x^2+5x-6\)

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

=x(x+6)-(x+6)

=(x+6)(x-1)

h: \(x^2-4x-5\)

\(=x^2-5x+x-5\)

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

=(x-5)(x+1)

i: \(16x-5x^2-3\)

\(=-5x^2+15x+x-3\)

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

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

j: \(x^2+x-20\)

\(=x^2+5x-4x-20\)

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

=(x+5)(x-4)

k: \(\left(x^2+9\right)^2-36x^2\)

\(=\left(x^2-6x+9\right)\left(x^2+6x+9\right)\)

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

l: \(x^6-x^4-4x^3+4x\)

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

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

3:

a: =>(x-2)(x+2)-3(x-2)^2=0

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

=>(x-2)(-2x+8)=0

=>x=2 hoặc x=4

b: =>15x+10=12x+1

=>3x=-9

=>x=-3

c; =>x-3=0 hoặc 2x+4=0

=>x=-2 hoặc x=3

d: =>2/3x-1-1/4x+1/4=3

=>5/12x-3/4=3

=>5/12x=15/4

=>x=9

2 should not go out for a walk

3 will quit it at once

4 must buy a smaller one

5 may be very worried

6 may find another one

7 can understand the lecture well

8 has to leave the building in order

9 ought to come to visit me more

10 must be very happy

Để A nguyên thì \(x-3\in\left\{-1;1;2;-2\right\}\)

hay \(x\in\left\{2;4;5;1\right\}\)

Đợi hết 5 phút đi r t giải cho =))

4: TH1: \(m^2-3<1\)

=>\(m^2<4\)

=>-2<m<2

=>A\(\cap\) B=∅

TH2: \(m^2-3=1\)

=>\(m^2=4\)

=>m=2 hoặc m=-2

A=(-∞;1]; B=[1;+∞)

A\(\cap\) B=(-∞;1]\(\cap\) [1;+∞)

={1}

TH3: \(m^2-3>1\)

=>\(m^2>4\)

=>m>2 hoặc m<-2

A=(-∞;\(m^2-3\) ]; B=[1;+∞)

mà \(m^2-3>1\)

nên A\(\cap\) B=[1;\(m^2-3\) ]

Bài 2:

a: A={x∈R| |x|>=2}

=>A=[2;+∞) \(\cup\) (-∞;-2]

|1-2x|<=4

=>|2x-1|<=4

=>-4<=2x-1<=4

=>-3<=2x<=5

=>\(-\frac32\le x\le\frac52\)

=>B=[-3/2;5/2]

b: A\(\cap\) B=([2;+∞) \(\cup\) (-∞;-2])\(\cap\) [-3/2;5/2]

=[2;5/2]

A\(\cup\) B=([2;+∞) \(\cup\) (-∞;-2])\(\cup\) [-3/2;5/2]

=R

A\B=([2;+∞) \(\cup\) (-∞;-2])\[-3/2;5/2]

=(-∞;-2]\(\cup\) (5/2;+∞)

B\A=[-3/2;5/2]\([2;+∞) \(\cup\) (-∞;-2])

=∅

\(C_{R}B\) =R\B=R\[-3/2;5/2]

=(-∞;-3/2)\(\cup\) (5/2;+∞)

b: C=(m;m+1)

A=[2;+∞) \(\cup\) (-∞;-2]

Để C⊂A thì m>=2 hoặc m+1<=-2

=>m>=2 hoặc m<=-3

tế bào là cấu trúc cơ bản của mọi cơ thể 

hình dạng và kích thước rất nhỏ 

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