ở oooo

e: \(\Leftrightarrow3x\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

a) \(215+x=400\)

\(\Rightarrow x=400-215\)

\(\Rightarrow x=185\)

b) \(12,5-2x=\dfrac{1}{5}\)

\(\Rightarrow2x=12,5-\dfrac{1}{5}\)

\(\Rightarrow2x=\dfrac{123}{10}\)

\(\Rightarrow x=\dfrac{123}{10}:2\)

\(\Rightarrow x=\dfrac{123}{20}\)

còn câu c d nhanh cậu ơi

Bạn xem lại đề phần e nha .

\(c)x.0,25-\dfrac{3}{25}:x=2\dfrac{1}{4}\)

\(x.0,25-\dfrac{25}{3}.x=\dfrac{9}{4}\)

\(x.\left(0,25-\dfrac{25}{3}\right)=\dfrac{9}{4}\)

\(x.\dfrac{-97}{12}=\dfrac{9}{4}\)

\(x=\dfrac{-27}{97}\)

\(d)\left(x-\dfrac{1}{3}\right).\left(x-\dfrac{1}{4}\right)=0\)

Do : 2 số đó nhân với nhau bằng 0 nên một trong 2 số là số 0.

Do đó :

\(Th1:x-\dfrac{1}{3}=0\)

\(x=\dfrac{1}{3}\)

\(Th2:x-\dfrac{1}{4}=0\)

\(x=\dfrac{1}{4}\)

\(a,\Leftrightarrow\left(5x+1\right)\left(x-4\right)-\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(5x+1-x\right)=0\\ \Leftrightarrow5x\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\\ b,\Leftrightarrow2x^2-10x-2x^2-3x=26\\ \Leftrightarrow-13x=26\\ \Leftrightarrow x=-2\\ c,\Leftrightarrow x^3+1-x^3+3x=15\\ \Leftrightarrow3x=14\\ \Leftrightarrow x=\dfrac{14}{3}\)

\(d,\Leftrightarrow x^3-5x+2x^2-10+5x-2x^2-17=0\\ \Leftrightarrow x^3-27=0\\ \Leftrightarrow x^3=27\\ \Leftrightarrow x=3\)

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

\(=x\left(x^2-16\right)-\left(x^4-1\right)\)

\(=x^3-16x-x^4+1\)

bạn ktra lại đề

b)  \(x^4+2x^3+5x^2+4x-12\)

\(=x^3\left(x-1\right)+3x^2\left(x-1\right)+8x\left(x-1\right)+12\left(x-1\right)\)

\(=\left(x-1\right)\left(x^3+3x^2+8x+12\right)\)

\(=\left(x-1\right)\left[x^2\left(x+2\right)+x\left(x+2\right)+6\left(x+2\right)\right]\)

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

Ủa pạn có thể giải ại cái bước thứ 2 đc ko ạk

giúp mình vs ạ

a) \(\left(2x-3\right)^2-\left(2x+5\right)^2=10\)

\(\Leftrightarrow4x^2-12x+9-4x^2-20x-25-10=0\)

\(\Leftrightarrow-32x-26=0\)

\(\Leftrightarrow-32x=26\)

\(\Rightarrow x=-\frac{13}{16}\)

b) \(4\left(x+1\right)^2+\left(2x-1\right)^2+8\left(x-1\right)\left(x+1\right)=11\)

\(\Leftrightarrow4x^2+8x+4+4x^2-4x+1+8x^2-8=0\)

\(\Leftrightarrow16x^2+4x-3=0\)

\(\Leftrightarrow4\left(4x^2+x+\frac{1}{16}\right)-\frac{13}{4}=0\)

\(\Leftrightarrow\left[2\left(2x+\frac{1}{4}\right)\right]^2-\left(\frac{\sqrt{13}}{2}\right)^2=0\)

\(\Leftrightarrow\left(4x+\frac{1}{2}-\frac{\sqrt{13}}{2}\right)\left(4x+\frac{1}{2}+\frac{\sqrt{13}}{2}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}4x+\frac{1-\sqrt{13}}{2}=0\\4x+\frac{1+\sqrt{13}}{2}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{\sqrt{13}-1}{8}\\x=\frac{-1-\sqrt{13}}{8}\end{cases}}\)

c) \(\left(x+5\right)^2=45+x^2\)

\(\Leftrightarrow x^2+10x+25-x^2-45=0\)

\(\Leftrightarrow10x-20=0\)

\(\Leftrightarrow10x=20\)

\(\Rightarrow x=2\)

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

\(\Leftrightarrow4x^2-12x+9-4x^2+4x-1+3=0\)

\(\Leftrightarrow-8x+11=0\)

\(\Leftrightarrow-8x=-11\)

\(\Rightarrow x=\frac{11}{8}\)

e) \(\left(x-1\right)^2-\left(5x-3\right)^2=0\)

\(\Leftrightarrow\left(x-1-5x+3\right)\left(x-1+5x-3\right)=0\)

\(\Leftrightarrow\left(-4x+2\right)\left(6x-4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}-4x+2=0\\6x-4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{2}{3}\end{cases}}\)

a) ( 2x - 3 )² - ( 2x + 5 )² = 10

<=> 4x² - 12x + 9 - ( 4x² + 20x + 25 ) = 10 

<=> 4x² - 12x + 9 - 4x² - 20x - 25 = 10

<=> -32x - 16 = 10

<=> -32x = 26

<=> x = -26/32 = -13/16

b) 4( x + 1 )² + ( 2x - 1 )² + 8( x + 1 )( x - 1 ) = 11

<=> 4( x² + 2x + 1 ) + ( 4x² - 4x + 1 ) + 8( x² - 1 ) = 11

<=> 4x² + 8x + 4 + 4x² - 4x + 1 + 8x² - 8 = 11

<=> 16x² + 4x - 14 = 0

<=> 2( 8x² + 2x - 7 ) = 0

<=> 8x² + 2x - 7 = 0

\(\Delta=b^2-4ac=2^2-4\cdot8\cdot\left(-7\right)=4+224=228\)( không muốn xài delta đâu nhưng bí quá ;-; )

\(\Delta>0\)nên phương trình đã cho có hai nghiệm phân biệt :

\(\hept{\begin{cases}x_1=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-2+\sqrt{228}}{2\cdot8}=\frac{-2+\sqrt{228}}{16}=\frac{-1+\sqrt{57}}{8}\\x_2=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-2-\sqrt{228}}{2\cdot8}=\frac{-2-\sqrt{228}}{18}=\frac{-1-\sqrt{57}}{8}\end{cases}}\)

( Mình nghĩ ý này bạn nên xem lại đề )

c) ( x + 5 )² = 45 + x²

<=> x² + 10x + 25 = 45 + x²

<=> x² + 10x - x² = 45 - 25

<=> 10x = 20

<=> x = 2 

d) ( 2x - 3 )² - ( 2x - 1 )² = -3

<=> 4x² - 12x + 9 - ( 4x² - 4x + 1 ) = -3

<=> 4x² - 12x + 9 - 4x² + 4x - 1 = -3

<=> -8x + 8 = -3

<=> -8x = -11

<=> x = 11/8

e) ( x - 1 )² - ( 5x - 3 )² = 0

<=> [ x - 1 - ( 5x - 3 ) ][ x - 1 + ( 5x - 3 ) ] = 0

<=> [ x - 1 - 5x + 3 ][ x - 1 + 5x - 3 ] = 0

<=> [ -4x + 2 ][ 6x - 4 ] = 0

<=> \(\orbr{\begin{cases}-4x+2=0\\6x-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{2}{3}\end{cases}}\)

a) 4(x+2) - 7(2x - 1) + 9(3x - 4)=30

⇔4x+8 - 14x + 7 + 27x - 36 = 30

⇔ 17x = 51

⇔ x = 3

b) 2(5x - 8) - 3(4x - 5) = 4(3x - 4) + 11

⇔ 10x - 16 - 12x + 15 = 12x - 16 + 11

⇔ -14x = -4

⇔ x= \(\frac{2}{7}\)

c) 5x(1 - 2x) - 3x(x + 18) = 0

⇔ 5x - 10x\(^2\) - 3x\(^2\) -54x =0

⇔ -13x\(^2\) -49 x = 0

⇔ -x ( 13x + 49 ) =0

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\13x+49=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{-49}{13}\end{matrix}\right.\)

d) 5x - 3{4x - 2[4x - 3(5x - 2)]} = 182

⇔ 5x - 3[ 4x - 2( 4x - 15x + 6 ) ]= 182

⇔5x - 3 ( 4x - 8x + 30x - 12 ) = 182

⇔ 5x - 3 ( 26x - 12 ) = 182

⇔ 5x - 78x + 36 = 182

⇔ - 73x = 146

⇔ x = -2

1/ 

a, (x-3)²+(4+x)(4-x)=10

<=>x²-6x+9+(16-x²)=10

<=>-6x+25=10

<=>-6x=-15

<=>x=5/2

còn lại tương tự a 

2/

a, \(a^2\left(a+1\right)+2a\left(a+1\right)=\left(a^2+2a\right)\left(a+1\right)=a\left(a+1\right)\left(a+2\right)\)

Vì a(a+1)(a+2) là tích 3 nguyên liên tiếp nên a(a+1)(a+2) chia hết cho 2,3

Mà (2,3)=1

=>a(a+1)(a+2) chia hết cho 6 (đpcm)

b, \(x^2+2x+2=\left(x^2+2x+1\right)+1=\left(x+1\right)^2+1\)

Vì \(\left(x+1\right)^2\ge0\Rightarrow\left(x+1\right)^2+1\ge1>0\left(đpcm\right)\)

c, \(x^2-x+1=\left(x^2-x+\frac{1}{4}\right)+\frac{3}{4}=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\)

Vì \(\left(x-\frac{1}{2}\right)^2\ge0\Rightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\)(đpcm)

d, \(-x^2+4x-5=-\left(x^2-4x+4\right)-1=-\left(x-2\right)^2-1\)

Vì \(-\left(x-2\right)^2\le0\Rightarrow-\left(x-2\right)^2-1\le-1< 0\) (đpcm)

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

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

\(\Leftrightarrow-4x^2+8x-4+4x^2-1=-3\)

\(\Leftrightarrow8x=2\)

\(\Leftrightarrow x=\frac{1}{4}\)

bn xem lại đi nha