x²-10x-9y²+25

=(x²-10x+25)-9y²

=(x-5)²-9y²

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

#H

x^2 -10x - 9y^2 +25 = x^2 -10x -9y^2 + 25 = -(3y-x+5)(3y+x-5)

Chúc bạn học tốt nha 

SAI

ĐỀ RỒI

BẠN 

ÊY

!!!!!

Trả lời:

Bài 1: 

a, \(A=\frac{x^2+x}{x^2-2x+1}:\left(\frac{x+1}{x}-\frac{1}{1-x}+\frac{2-x^2}{x^2-x}\right)\)\(\left(ĐKXĐ:x\ne0;x\ne1\right)\)

\(=\frac{x\left(x+1\right)}{\left(x-1\right)^2}:\left[\frac{x+1}{x}+\frac{1}{x-1}+\frac{2-x^2}{x\left(x-1\right)}\right]\)

\(=\frac{x\left(x+1\right)}{\left(x-1\right)^2}:\left[\frac{\left(x+1\right)\left(x-1\right)}{x\left(x-1\right)}+\frac{x}{x\left(x-1\right)}+\frac{2-x^2}{x\left(x-1\right)}\right]\)

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

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

\(=\frac{x\left(x+1\right)}{\left(x-1\right)^2}:\frac{x+1}{x\left(x-1\right)}=\frac{x\left(x+1\right)}{\left(x-1\right)^2}\cdot\frac{x\left(x-1\right)}{x+1}=\frac{x^2}{x-1}\)

b, \(\left|2x-5\right|=3\)

\(\Leftrightarrow\orbr{\begin{cases}2x-5=3\\2x-5=-3\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=8\\2x=2\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=4\left(tm\right)\\x=1\left(ktm\right)\end{cases}}}\)

Thay x = 4 vào A, ta có:

\(A=\frac{4^2}{4-1}=\frac{16}{3}\)

c, \(A=4\)

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

\(\Leftrightarrow\frac{x^2}{x-1}=\frac{4\left(x-1\right)}{x-1}\)

\(\Rightarrow x^2=4x-4\)

\(\Leftrightarrow x^2-4x+4=0\)

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

\(\Leftrightarrow x-2=0\)

\(\Leftrightarrow x=2\)

Vậy x = 2 thì A = 4

d, \(A< 2\)

\(\Leftrightarrow\frac{x^2}{x-1}< 2\)

\(\Leftrightarrow\frac{x^2}{x-1}-2< 0\)

\(\Leftrightarrow\frac{x^2-2\left(x-1\right)}{x-1}< 0\)

\(\Leftrightarrow\frac{x^2-2x+2}{x-1}< 0\)

\(\Leftrightarrow x-1< 0\) ( vì \(x^2-2x+2=\left(x^2-2x+1\right)+1=\left(x-1\right)^2+1\ge1>0\forall x\) )

\(\Leftrightarrow x< 1\)

Vậy x < 1 thì A < 2

( x² + 1 )² - 4x² = ( x² + 1 )² - ( 2x )² = ( x² - 2x + 1 )( x² + 2x + 1 ) = ( x - 1 )²( x + 1 )²

[x mũ 2 +1]^2 - 4x^2 = (x^2 + 1)^2 -4x^2 = (x-1)^2(x+1)^2

x³ - 19x - 30 = 0

<=> x³ - 5x² + 5x² - 25x + 6x - 30 = 0

<=> x²( x - 5 ) + 5x( x - 5 ) + 6( x - 5 ) = 0 

<=> ( x - 5 )( x² + 5x + 6 ) = 0

<=> ( x - 5 )( x² + 3x + 2x + 6 ) = 0

<=> ( x - 5 )[ x( x + 3 ) + 2( x + 3 ) ] = 0

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

đến đây dễ rồi :)

\(x^3-19x-30=0\)

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

\(\Leftrightarrow\hept{\begin{cases}x-5=0\\x+2=0\\x+3=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=5\\x=-2\\x=-3\end{cases}}}\)

Vậy B=x1²+x2²+x3²

B=5²+(-2)²+(-3)²

B=25+4+9

B=38

#H

3x^2 - 5x - 2 = 0

(a = 3; b = -5; c = -2)

ta có x1, x2 là nghiệm của pt nên : x1 + x2 = -b/a = -(-5)/3 = 5/3

vậy_

Áp dụng hệ thức Vi - ét ta có : x₁ + x₂ = \(-\frac{b}{a}=-\frac{-5}{3}=\frac{5}{3}\)

Hoặc chưa học ở lớp 8 thì \(3x^2-5x-2=3x^2-6x+x-2=3x\left(x-2\right)+\left(x-2\right)\)

\(=\left(3x+1\right)\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}x1=-\frac{1}{3}\\x2=2\end{cases}}\Leftrightarrow x1+x2=-\frac{1}{3}+2=\frac{5}{3}\)

là số 192 nha bạn 

chúc bạn học tốt 

số 192 nha bn

\(3x^2-5x-2=0\)

\(\Leftrightarrow3x^2-6x+x-2=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=2\end{cases}}\)

\(A=x_1+x_2=\frac{1}{3}+2=\frac{7}{3}\).

Câu 9. 

\(x^3-19x-30=0\)

\(\Leftrightarrow x^3+2x^2-2x^2-4x-15x-30=0\)

\(\Leftrightarrow\left(x+2\right)\left(x^2-2x-15\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x^2-5x+3x-15\right)=0\)

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

\(\Leftrightarrow\hept{\begin{cases}x_1=-2\\x_2=5\\x_3=-3\end{cases}}\)

\(B=x_1^2+x_2^2+x_3^2=38\)

Câu 10. 

\(2a>b>0\Rightarrow2\frac{a}{b}>1\Leftrightarrow t>\frac{1}{2}\)(\(t=\frac{a}{b}\))

\(4a^2+b^2=5ab\)

\(\Leftrightarrow4\left(\frac{a}{b}\right)^2+1-5\frac{a}{b}=0\)

\(\Leftrightarrow4t^2-4t-t+1=0\)

\(\Leftrightarrow\left(4t-1\right)\left(t-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}t=\frac{1}{4}\left(l\right)\\t=1\left(tm\right)\end{cases}}\)

\(A=\frac{ab}{4a^2-b^2}=\frac{t}{4t^2-1}=\frac{1}{4-1}=\frac{1}{3}\)