\(1.\text{ }\text{ }\text{ }\dfrac{\left(x^2+2\right)^2-4x^2}{y\left(x^2+2\right)-2xy-\left(x-1\right)^2-1}\\ =\dfrac{\left(x^2+2-2x\right)\left(x^2+2+2x\right)}{x^2y+2y-2xy-x^2+2x-1-1}\\ =\dfrac{\left(x^2+2-2x\right)\left(x^2+2+2x\right)}{\left(x^2y-x^2\right)-\left(2xy-2x\right)+\left(2y-2\right)}\\ =\dfrac{\left(x^2+2-2x\right)\left(x^2+2+2x\right)}{x^2\left(y-1\right)-2x\left(y-1\right)+2\left(y-1\right)}\\ =\dfrac{\left(x^2+2-2x\right)\left(x^2+2+2x\right)}{\left(x^2-2x+2\right)\left(y-1\right)}\\ =\dfrac{x^2+2x+2}{y-1}\)

\(2.\text{ }\text{ }\text{ }\text{ }\dfrac{x^2+5x+6}{x^2+3x+2}\\ =\dfrac{x^2+3x+2x+6}{x^2+2x+x+2}\\ =\dfrac{\left(x^2+3x\right)+\left(2x+6\right)}{\left(x^2+2x\right)+\left(x+2\right)}\\ =\dfrac{x\left(x+3\right)+2\left(x+3\right)}{x\left(x+2\right)+\left(x+2\right)}\\ =\dfrac{\left(x+2\right)\left(x+3\right)}{\left(x+2\right)\left(x+1\right)}\\ =\dfrac{x+3}{x+1}\)

\(3.\text{ }\text{ }\text{ }\dfrac{x^2+y^2-z^2-2zt+2xy-t^2}{x^2-y^2+z^2-2yt+2xz-t^2}\text{ ( Chữa đề ) }\\ =\dfrac{\left(x^2+2xy+y^2\right)-\left(z^2+2zt+t^2\right)}{\left(x^2+2xz+z^2\right)-\left(y^2+2yt+t^2\right)}\\ =\dfrac{\left(x+y\right)^2-\left(z+t\right)^2}{\left(x+z\right)^2-\left(y+t\right)^2}\\ =\dfrac{\left(x+y+z+t\right)\left(x+y-z-t\right)}{\left(x+z+y+t\right)\left(x+z-y-t\right)}\\ =\dfrac{x+y-z-t}{x+z-y-t}\)

\(4.\text{ }\text{ }\text{ }\dfrac{\left(n+1\right)!}{\left(n+1\right)!+\left(n+2\right)!}=\dfrac{\left(n+1\right)!}{\left(n+1\right)!\left(1+n+2\right)}=\dfrac{1}{n+3}\)

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

\(6.\text{ }\text{ }\text{ }\dfrac{x^2-3x}{2x^2-7x+3}\\ =\dfrac{x\left(x-3\right)}{2x^2-6x-x+3}\\ =\dfrac{x\left(x-3\right)}{\left(2x^2-6x\right)-\left(x-3\right)}\\ =\dfrac{x\left(x-3\right)}{2x\left(x-3\right)-\left(x-3\right)}\\ =\dfrac{x\left(x-3\right)}{\left(2x-1\right)\left(x-3\right)}\\ =\dfrac{x}{2x-1}\)

a) \(A = \frac{2x^2 - 16x+43}{x^2-8x+22}\) = \(\frac{2(x^2-8x+22)-1}{x^2-8x+22}\) = \(2 - \frac{1}{x^2-8x+22}\)

Ta có : \(x^2-8x+22 \) = \(x^2-8x+16+6 = ( x-4)^2 +6 \)

Vì \((x-4)^2 \ge 0 \) với \( \forall x\in R\) Nên \(( x-4)^2 +6 \ge 6 \)

\(\Rightarrow \) \(x^2-8x+22 \) \( \ge 6\)\(\Rightarrow \) \(\frac{1}{x^2-8x+22} \) \(\le \frac{1}{6}\) \(\Rightarrow \) - \(\frac{1}{x^2-8x+22} \) \(\ge - \frac{1}{6}\)

\(\Rightarrow \) A = \(2 - \frac{1}{x^2-8x+22}\) \( \ge 2-\frac{1}{6}\) = \(\frac{11}{6}\) Dấu "=" xảy ra khi và chỉ khi x=4

Vậy GTNN của A = \(\frac{11}{6}\) khi và chỉ khi x=4

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

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

b: |2x+1|=5

=>2x+1=5 hoặc 2x+1=-5

=>2x=4 hoặc 2x=-6

=>x=2(nhận) hoặc x=-3(nhận)

Khi x=2 thì \(M=\dfrac{4+6-3}{\left(2+1\right)\left(2^2-1\right)}=\dfrac{7}{3\cdot3}=\dfrac{7}{9}\)

Khi x=-3 thì \(M=\dfrac{9-9-3}{\left(-3+1\right)\left(9-1\right)}=\dfrac{-3}{\left(-2\right)\cdot8}=\dfrac{3}{16}\)

Bài 1 . Chia :( x³ + 5x² - 4x - 20) cho ( x² + 3x - 10) ta được x+ 2

Chia :( x³ + 5x² - 4x - 20) cho ( x² + 7x + 10) ta được x - 2

Do đó , ta có :

\(\dfrac{1}{x^2+3x-10}=\dfrac{x+2}{\left(x^2+3x-10\right)\left(x+2\right)}=\dfrac{x+2}{x^3+5x^2-4x-20}\)

Và : \(\dfrac{x}{x^2+7x+10}=\dfrac{x\left(x-2\right)}{\left(x^2+7x+10\right)\left(x-2\right)}=\dfrac{x^2-2x}{x^3+5x^2-4x-20}\)

Bài 2 . a) Ta có :

\(\dfrac{x-1}{x^3+1}\)( giữ nguyên)

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

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

b) Ta có MTC = x²( y - z)²

Ta có :

\(\dfrac{x+y}{x\left(y-z\right)^2}=\dfrac{x^2+xy}{x^2\left(y-z\right)^2}\)

\(\dfrac{y}{x^2\left(y-z\right)^2}\)( giữ nguyên )

\(\dfrac{z}{x^2}=\dfrac{z\left(y-z\right)^2}{x^2\left(y-z\right)^2}\)

bài này phần luyện tập à bạn

A= \(\dfrac{3x+2}{x-3}\)= \(\dfrac{3\left(x-3\right)+11}{x-3}\)= 3 + \(\dfrac{11}{x-3}\)

Để A là số nguyên <=> \(\dfrac{11}{x-3}\) là số nguyên

<=> 11 chia hết cho x-3

<=> x-3 thuộc Ư(11)

Ta có bảng sau

Vậy x thuộc { 4;2;14;-8}

a, A= \(\dfrac{3x+2}{x-3}\)

Để A là số nguyên⇒ 3x+ 2⋮ x- 3

Vì x- 3⋮ x- 3

⇒ 3.(x- 3)⋮ x- 3

⇒ 3x- 3.3⋮ x-3

⇒ 3x- 9⋮ x-3

Mà 3x+ 2⋮ x-3

⇒ ( 3x+ 2)- ( 3x- 9)⋮ x-3

⇒ 3x+ 2- 3x+ 9⋮ x-3

⇒ ( 3x- 3x)+ ( 2+ 9)⋮ x- 3

⇒ 11⋮ x- 3

⇒ x- 3∈ Ư(11)

⇒ x- 3∈ ( -11; -1; 1; 11)

⇒ x∈ ( -8; 2; 4; 14)

Vậy....................

b, B= \(\dfrac{x^2+3x-7}{x+3}\)

Để B là số nguyên⇒ x²+3x-7 ⋮ x+3

Vì x+ 3⋮ x+ 3

⇒ x(x+3)⋮ x+ 3

⇒ x²+x.3⋮ x+ 3

Mà x²+ 3x- 7⋮ x+ 3

⇒ (x²+x.3)-( x²+3x-7)⋮ x+ 3

⇒ x²+ x.3- x² -3x+ 7⋮ x+3

⇒ (x²-x²)+(3x- 3x)+ 7⋮ x+ 7

⇒ 7⋮ x+ 7

⇒ x+ 7∈ Ư(7)

⇒ x+ 7∈ (-7; -1; 1; 7)

⇒ x∈ ( -14; -8; -6; 0)

Vậy......................................

c, C= \(\dfrac{2x-1}{x+2}\)

Để C là số nguyên⇒ 2x-1⋮ x+2

Vì x+ 2⋮ x+2

⇒ 2( x+2)⋮ x+2

⇒ 2x+ 4⋮ x+2

Mà 2x- 1⋮ x+2

⇒ (2x+4)- (2x-1)⋮ x+2

⇒ 2x+ 4- 2x+ 1⋮ x+2

⇒ (2x-2x)+ (4+1)⋮ x+2

⇒ 5⋮ x+2

⇒ x+2∈ Ư(5)

⇒ x+2∈ (-5; -1; 1; 5)

⇒ x∈ ( -7; -3; -1; 3)

Vậy..........................................

Sửa đề nhé\(\dfrac{1}{3x+3y+2z}=\dfrac{1}{\left(z+x\right)+\left(z+y\right)+\left(x+y\right)+\left(x+y\right)}\)

\(\le\dfrac{1}{16}\left(\dfrac{1}{x+z}+\dfrac{1}{z+y}+\dfrac{1}{x+y}+\dfrac{1}{x+y}\right)\)

CMTT và cộng theo vế:

\(VT\le\dfrac{1}{16}\left(\dfrac{1}{x+z}+\dfrac{1}{z+y}+\dfrac{1}{x+y}+\dfrac{1}{x+y}+\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{x+z}+\dfrac{1}{x+z}+\dfrac{1}{x+z}+\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{y+z}\right)\)

\(=\dfrac{1}{16}.24=\dfrac{3}{2}\)

\("="\Leftrightarrow x=y=z=\dfrac{1}{4}\)

đúng rùi đó

Sai

Lời giải:
a)

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

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

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

b)

Xét tử số:

\(x^3+y^3+z^3-3xyz=(x+y)^3-3xy(x+y)+z^3-3xyz\)

\(=[(x+y)^3+z^3]-3xy(x+y+z)\)

\(=(x+y+z)[(x+y)^2-(x+y)z+z^2]-3xy(x+y+z)\)

\(=(x+y+z)[(x+y)^2-(x+y)z+z^2-3xy]\)

\(=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)\)

Do đó:

\(\frac{x^3+y^3+z^3-3xyz}{x^2+y^2+z^2-xy-yz-xz}=\frac{(x+y+z)(x^2+y^2+z^2-xy-yz-xz)}{x^2+y^2+z^2-xy-yz-xz}=x+y+z\)

1.

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

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

=> \(\dfrac{1}{x+y+z}\) = 2

=> x+y+z = \(\dfrac{1}{2}\)

Ta có: \(\dfrac{y+z+1}{x}\) = 2

=> y+z+1 = 2x => x+y+z+1 = 3x <=> \(\dfrac{3}{2}=3x\)

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

Tương tự thế vào \(\dfrac{x+z+2}{y}\) tính được y =\(\dfrac{5}{6}\)

=> z = -\(\dfrac{5}{6}\)

=> A = 2016.\(\dfrac{1}{2}\) = 1008

Bài 1: 

a: \(A=\dfrac{x^4+x^3+x+1}{x^4-x^3+2x^2-x+1}=\dfrac{x^3\left(x+1\right)+\left(x+1\right)}{x^4-x^3+x^2+x^2-x+1}\)

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

Để A=0 thì x+1=0

hay x=-1

b: \(B=\dfrac{x^4-5x^2+4}{x^4-10x^2+9}=\dfrac{\left(x^2-1\right)\left(x^2-4\right)}{\left(x^2-1\right)\left(x^2-9\right)}=\dfrac{x^2-4}{x^2-9}\)

Để B=0 thi (x-2)(x+2)=0

=>x=2 hoặc x=-2