e)(3x-1)(2x+7)-(x+1)(6x-5)=16

=>\(6x^2-2x+21x-7-6x^2-6x+5x+5\)=16

=>18x-2=16

=> 18x=18

=> x=1

b)\(x\left(x+1\right)\left(x+6\right)-x^3=5x\)

=>\(x\left(x+1\right)\left(x+6\right)-x^3-5x=0\)

=> \(x\left(x^2+x+6x+6\right)-x\left(x^2-5\right)=0\)

=>\(x\left[\left(x^2+7x+6\right)-\left(x^2-5\right)\right]=0\)

=> __x=1

     |__7x+1=0=> 7x=-1=> x=-1/7

c)3(2x-1)-5(x-3)+6(3x-4)=24

<=>6x-3-5x-15+18x-24=24

<=>19x-12=24

<=>19x=36

<=>x=\(\frac{36}{19}\)

d)2x(5-3x)+2x(3x-5)-3(x-7)=3

<=>10x-6x²+6x²-10x-3x-21=3

<=>-3(x-7)=3

<=>21-3x=3

<=>-3x=-18

<=>x=6

Ai trả lời câu này giúp em và nhỏ Vi với

a.\(6x^2-\left(2x-3\right)\left(3x+2\right)-1=0\Leftrightarrow6x^2-\left(6x^2-2x-6\right)-1=0\)

\(\Leftrightarrow2x+5=0\Leftrightarrow x=-\frac{5}{2}\)

b. \(\left(x-3\right)\left(x+7\right)-\left(x+5\right)\left(x-1\right)=0\Leftrightarrow x^2+4x-21-\left(x^2+4x-5\right)=0\)

\(\Leftrightarrow-16=0\)

Vậy không có x thỏa mãn.

Bài 1

a) 5x²y - 20xy²

= 5xy(x - 4y)

b) 1 - 8x + 16x² - y²

= (1 - 8x + 16x²) - y²

= (1 - 4x)² - y²

= (1 - 4x - y)(1 - 4x + y)

c) 4x - 4 - x²

= -(x² - 4x + 4)

= -(x - 2)²

d) x³ - 2x² + x - xy²

= x(x² - 2x + 1 - y²)

= x[(x² - 2x+ 1) - y²]

= x[(x - 1)² - y²]

= x(x - 1 - y)(x - 1 + y)

= x(x - y - 1)(x + y - 1)

e) 27 - 3x²

= 3(9 - x²)

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

f) 2x² + 4x + 2 - 2y²

= 2(x² + 2x + 1 - y²)

= 2[(x² + 2x + 1) - y²]

= 2[(x + 1)² - y²]

= 2(x + 1 - y)(x + 1 + y)

= 2(x - y + 1)(x + y + 1)

Bài 2:

a: \(x^2\left(x-2023\right)+x-2023=0\)

=>\(\left(x-2023\right)\left(x^2+1\right)=0\)

mà \(x^2+1>=1>0\forall x\)

nên x-2023=0

=>x=2023

b: 

ĐKXĐ: x<>0

\(-x\left(x-4\right)+\left(2x^3-4x^2-9x\right):x=0\)

=>\(-x\left(x-4\right)+2x^2-4x-9=0\)

=>\(-x^2+4x+2x^2-4x-9=0\)

=>\(x^2-9=0\)

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

=>\(\left[{}\begin{matrix}x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

c: \(x^2+2x-3x-6=0\)

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

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

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

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

d: 3x(x-10)-2x+20=0

=>\(3x\left(x-10\right)-\left(2x-20\right)=0\)

=>\(3x\left(x-10\right)-2\left(x-10\right)=0\)

=>\(\left(x-10\right)\left(3x-2\right)=0\)

=>\(\left[{}\begin{matrix}x-10=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=10\end{matrix}\right.\)

Câu 1:

a: \(5x^2y-20xy^2\)

\(=5xy\cdot x-5xy\cdot4y\)

\(=5xy\left(x-4y\right)\)

b: \(1-8x+16x^2-y^2\)

\(=\left(16x^2-8x+1\right)-y^2\)

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

\(=\left(4x-1-y\right)\left(4x-1+y\right)\)

c: \(4x-4-x^2\)

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

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

d: \(x^3-2x^2+x-xy^2\)

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

\(=x\left[\left(x^2-2x+1\right)-y^2\right]\)

\(=x\left[\left(x-1\right)^2-y^2\right]\)

\(=x\left(x-1-y\right)\left(x-1+y\right)\)

e: \(27-3x^2\)

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

\(=3\left(3-x\right)\left(3+x\right)\)

f: \(2x^2+4x+2-2y^2\)

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

\(=2\left[\left(x^2+2x+1\right)-y^2\right]\)

\(=2\left[\left(x+1\right)^2-y^2\right]\)

\(=2\left(x+1+y\right)\left(x+1-y\right)\)

Bài 2

a) x²(x - 2023) - 2023 + x = 0

x²(x - 2023) - (x - 2023) = 0

(x - 2023)(x² - 1) = 0

x - 2023 = 0 hoặc x² - 1 = 0

*) x - 2023 = 0

x = 2023

*) x² - 1 = 0

x² = 1

x = 1 hoặc x = -1

Vậy x = -1; x = 1; x = 2023

b) -x(x - 4) + (2x³ - 4x² - 9x) : x = 0

-x² + 4x + 2x² - 4x - 9 = 0

x² - 9 = 0

x² = 9

x = 3 hoặc x = -3

Vậy x = 3; x = -3

c) x² + 2x - 3x - 6 = 0

(x² + 2x) - (3x + 6) = 0

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

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

x + 2 = 0 hoặc x - 3 = 0

*) x + 2 = 0

x = -2

*) x - 3 = 0

x = 3

Vậy x = -2; x = 3

d) 3x(x - 10) - 2x + 20 = 0

3x(x - 10) - (2x - 20) = 0

3x(x - 10) - 2(x - 10) = 0

(x - 10)(3x - 2) = 0

x - 10 = 0 hoặc 3x - 2 = 0

*) x - 10 = 0

x = 10

*) 3x - 2 = 0

3x = 2

x = 2/3

Vậy x = 2/3; x = 10

\(\left(\frac{2}{3}.x+\frac{5}{6}\right).\left(\frac{3}{4}.x-\frac{27}{8}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}\frac{2}{3}.x+\frac{5}{6}=0\\\frac{3}{4}.x-\frac{27}{8}=0\end{cases}}\Rightarrow\orbr{\begin{cases}\frac{2}{3}.x=\frac{-5}{6}\\\frac{3}{4}.x=\frac{27}{8}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{-5}{4}\\x=\frac{9}{2}\end{cases}}\)

Vậy \(x=\frac{-5}{4}\) hoặc \(x=\frac{9}{2}\).

Học tốt

\(\left(\frac{2}{3}.x+\frac{5}{6}\right).\left(\frac{3}{4}.x-\frac{27}{8}\right)=0=>\orbr{\begin{cases}\left(\frac{3}{4}.x-\frac{27}{8}\right)=0\\\left(\frac{2}{3}.x+\frac{5}{6}\right)=0\end{cases}}\)

\(\orbr{\begin{cases}\frac{2}{3}.x=0-\frac{5}{6}\\\frac{3}{4}.x=0+\frac{27}{8}\end{cases}}=>\orbr{\begin{cases}\frac{2}{3}.x=-\frac{5}{6}\\\frac{3}{4}.x=\frac{27}{8}\end{cases}}=>\orbr{\begin{cases}x=-\frac{5}{6}:\frac{2}{3}\\x=\frac{27}{8}:\frac{3}{4}\end{cases}}=>\orbr{\begin{cases}x=-\frac{5}{4}\\x=\frac{9}{2}\end{cases}}\)

vậy \(\orbr{\begin{cases}x=-\frac{5}{4}\\x=\frac{9}{2}\end{cases}}\)

                Bài làm :

\(\left(\frac{2}{3}x+\frac{5}{6}\right).\left(\frac{3}{4}x-\frac{27}{8}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{2}{3}x+\frac{5}{6}=0\\\frac{3}{4}x-\frac{27}{8}=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{2}{3}x=\frac{-5}{6}\\\frac{3}{4}x=\frac{27}{8}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-5}{4}\\x=\frac{9}{2}\end{cases}}\)

Vậy x=-5/4 hoặc x=9/2

1. ĐKXĐ: ...

Do \(\sqrt{x-2}\ge0\Rightarrow\frac{3}{\sqrt{x-2}+3}\le1\)

\(\frac{1}{\sqrt{x+6}+3}>0\)

\(\Rightarrow VT< 1-0=1< 2\)

Pt vô nghiệm

2.

ĐKXĐ: \(x\ge\frac{1}{2}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+2x}=a>0\\\sqrt{2x-1}=b\ge0\end{matrix}\right.\)ta được:

\(a+b=\sqrt{3a^2-b^2}\)

\(\Leftrightarrow a^2+b^2+2ab=3a^2-b^2\)

\(\Leftrightarrow a^2-ab-b^2=0\) (nghiệm xấu quá bạn coi lại đề)

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

\(\Leftrightarrow a=\frac{\sqrt{5}+1}{2}b\)

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

\(\Leftrightarrow x^2-2\left(\frac{1+\sqrt{5}}{2}\right)x+\frac{3+\sqrt{5}}{2}=0\)

4. Bạn coi lại đề (chính xác là pt này ko có nghiệm thực)

5.

\(\Leftrightarrow x^2+x+6-\left(2x+1\right)\sqrt{x^2+x+6}+6x-6=0\)

Đặt \(\sqrt{x^2+x+6}=t>0\)

\(t^2-\left(2x+1\right)t+6x-6=0\)

\(\Delta=\left(2x+1\right)^2-4\left(6x-6\right)=\left(2x-5\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}t=\frac{2x+1+2x-5}{2}=2x-2\\t=\frac{2x+1-2x+5}{2}=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+x+6}=2x-2\left(x\ge1\right)\\\sqrt{x^2+x+6}=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+x+6=4x^2-8x+4\left(x\ge1\right)\\x^2+x+6=9\end{matrix}\right.\)

6.

Đặt \(\left\{{}\begin{matrix}\sqrt{5x^2+6x+5}=a\\4x=b\end{matrix}\right.\)

\(\Rightarrow a\left(a^2+1\right)=b\left(b^2+1\right)\)

\(\Leftrightarrow a^3-b^3+a-b=0\)

\(\Leftrightarrow\left(a-b\right)\left(a^2+b^2+ab+1\right)=0\)

\(\Leftrightarrow a=b\)

\(\Leftrightarrow\sqrt{5x^2+6x+5}=4x\left(x\ge0\right)\)

\(\Leftrightarrow5x^2+6x+5=16x^2\)

\(\Leftrightarrow11x^2-6x-5=0\)

\(\Rightarrow x=1\)

\(f\left(x\right)=\left(x-1\right).g\left(x\right)\)

\(\Rightarrow3x^3-2x^2+x+5=\left(x-1\right)\left(3x^2+ax+b\right)\)

\(\Rightarrow3x^3-2x^2+x+5=3x^3+ax^2+bx-3x^2-ax-b\)

\(\Rightarrow-2x^2+x+5=x^2\left(a-3\right)+x\left(b-a\right)-b\)

-Bạn kiểm tra lại đề.