\(a,\Leftrightarrow x^2-2x-x^2+1=0\\ \Leftrightarrow-2x+1=0\Leftrightarrow x=\dfrac{1}{2}\\ b,\Leftrightarrow\left(2x-1-x-4\right)\left(2x-1+x+4\right)=0\\ \Leftrightarrow\left(x-5\right)\left(3x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)

\(1,\\ a,\Leftrightarrow x-\dfrac{1}{3}=0\Leftrightarrow x=\dfrac{1}{3}\\ b,\Leftrightarrow\left[{}\begin{matrix}x-4=4\\x-4=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=0\end{matrix}\right.\\ c,\Leftrightarrow2x+1=-2\Leftrightarrow x=-\dfrac{3}{2}\\ 2,\\ a,=1\\ b,=\left(\dfrac{13}{4}\right)^2=\dfrac{169}{16}\\ c,=\left(-\dfrac{7}{4}\right)^2=\dfrac{49}{16}\\ d,=\left(\dfrac{3}{7}\right)^{20}:\left(\dfrac{3}{7}\right)^{12}=\left(\dfrac{3}{7}\right)^8=...\\ e,=\left(3\cdot5\cdot\dfrac{2}{3}\right)^2=10^2=100\)

cần gấp lắm ạ xin các bạn giúp đỡ . ___ .

Câu 1: 

0,9 x 218 x 2 + 0,18 x 4290 + 0,6 x 353 x 3

= 9/10 x 436 + 9/50 x 4290 + 6/10 x 1059

= 9 x 43,6 + 9 x 85,8 + 6 x 105,9

= 3 x 130,8 + 3 x 257,4 + 3 x 211,8

= 3 x ( 130,8 + 257,4 + 211,8 )

= 3 x 600 

= 1800

Câu 2: 

3/4 x X + 1/2 x X - 15 = 35

X x ( 3/4 + 1/2 ) - 15 = 35

X x ( 3/4 + 1/2 ) = 50

X x 5/4 = 50

X = 40

VẬy X = 40

1.  0.9x218x2+0.18x4290+0.6x353x3  

= 1.8x218+0.18x10x429+1.8x353

=1.8x218+1.8x429+1.8x353

= 1.8x( 218+429+353 )

= 1.8x1000

=1800

2. 

3/4 x X + 1/2 x X - 15 = 35

3/4 x X + 1/2 x X = 35+15

3/4 x X +1/2 x X = 50

X x ( 3/4 + 1/2 ) = 50 

X x 5/4 = 50 

X = 50 : 5/4

X =40

1) Ta có: |x+3| \(\ge\)0; |2x+y-4| \(\ge\)0

\(\Rightarrow\) |x + 3| + |2x + y - 4| \(\ge\) 0

Dấu = xảy ra khi x+3=0 và 2x+y-4 = 0 \(\Rightarrow\)x=-3; y=10

1)  |x + 3| + |2x + y - 4| = 0

\(\Leftrightarrow\hept{\begin{cases}x+3=0\\2x+y-4=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-3\\-6+y-4=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-3\\y=10\end{cases}}\)

2) Ta có:  |x + 3| + |x + y - 6| + |7 - 1| = 0

\(\Leftrightarrow\) |x + 3| + |x + y - 6| + 6= 0

\(\Leftrightarrow\)|x + 3| + |x + y - 6| = -6

a :

[( 10 - x ) . 2 + 5] : 3 - 2 = 3

( 10 - x ) x 2 + 5 = ( 3+2) x 3

( 10 - x ) x 2 + 5 = 15

10 -x  = ( 15 -5 ) : 2

10 - x = 5

x = 5

a) \(\left[\left(10-x\right)2+5\right]:3-2=3\)

\(\left(20-2x+5\right):3-2=3\)

\(\left(20-2x+5\right):3=5\)

\(20-2x+5=15\)

\(20-2x=10\)

\(2x=10\)

\(x=5\)

b) \(6x-302=2^3\cdot5\)

\(6x=8\cdot5+302\)

\(6x=342\)

\(x=57\)

c) \(12\left(x-1\right):3=4^3-2^3\)

\(12\left(x-1\right):3=56\)

\(12\left(x-1\right)=168\)

\(x-1=12\)

\(x=13\)

a, [(10-x).2+5]:3-2=3

    [(10-x).2+5]:3  =3+2

    [(10-x).2+5]:3  =5

    (10-x).2+5       =5.3

    (10-x).2+5       =15

    (10-x).2           =15-5

    (10-x).2           =10

    10-x                =10:2

    10-x                =5

        x                =10-5

        x                =5

b, 6x-320=2³.5

    6x-320=8.5

    6x-320=40

    6x      =40+320

    6x      =360

     x       =360:6

     x       =60

c, 12.(x-1):3=4³-2³

    12.(x-1):3=64-8

    12.(x-1):3=56

    12.(x-1)   =56.3

    12.(x-1)   =168

          x-1   =168:12

          x-1   =14

          x      =14+1

          x      =15

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x^2-x+6=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\\left[x^2-2\cdot x\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right]+\dfrac{23}{4}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\left(N\right)\\\left(x-\dfrac{1}{2}\right)^2+\dfrac{23}{4}\ge\dfrac{23}{4}>0\left(L\right)\end{matrix}\right.\)

Vậy \(S=\left\{-2\right\}\)

b) \(9x^2-4-\left(3x-2\right)^2=0\)

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

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

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

\(\Leftrightarrow\left(3x-2\right)\cdot4=0\)

\(\Leftrightarrow3x-2=0\)

\(\Leftrightarrow x=\dfrac{2}{3}\)

Vậy \(S=\left\{\dfrac{2}{3}\right\}\)

`@` `\text {Ans}`

`\downarrow`

`a)`

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

`=> 5*x^3 = 0+5`

`=> 5*x^3 = 5`

`=> x^3 = 5 \div 5`

`=> x^3 = 1`

`=> x^3 = 1^3`

`=> x=1`

Vậy, `x=1.`

`b)`

\(( x+1)^2 = 16\)

`=> (x+1)^2 = (+-4)^2`

`=>`\(\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=4-1\\x=-4-1\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)

Vậy, `x \in {3; -5}`

`c)`

\(( x+1)^3 = 27\)

`=> (x+1)^3 = 3^3`

`=> x+1=3`

`=> x=3-1`

`=> x=2`

Vậy, `x=2.`

`d)`

\(( x-1)^3 = 343\)

`=> (x-1)^3 = 7^3`

`=> x-1=7`

`=> x=7+1`

`=> x=8`

Vậy, `x=8.`

`e)`

\((2x - 1^3) = 125\) hay đề là `(2x-1)^3 = 125` vậy ạ?

Mình làm cả 2 TH nhé!

`(2x-1^3)=125`

`=> 2x-1=125`

`=> 2x=125+1`

`=> 2x=126`

`=> x=126 \div 2`

`=> x=63`

TH2:

`(2x-1)^3 = 125`

`=> (2x-1)^3 = 5^3`

`=> 2x-1=5`

`=> 2x=5+1`

`=> 2x=6`

`=> x=6 \div 2`

`=> x=3`

Vậy, `x=3.`

(a) \(5x^3-5=0\Leftrightarrow5x^3=5\Leftrightarrow x^3=1\Leftrightarrow x=1\)

(b) \(\left(x+1\right)^2=16\Rightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)

(c) \(\left(x+1\right)^3=27\Leftrightarrow x+1=3\Leftrightarrow x=2\)

(d) \(\left(x-1\right)^3=343\Leftrightarrow x-1=7\Leftrightarrow x=8\)

(e) \(\left(2x-1\right)^3=125\Leftrightarrow2x-1=5\Leftrightarrow2x=6\Leftrightarrow x=3\)

a. (2x + 1)² - 4x² + 2x² - 2 = 0

<=> (2x + 1 - 2x)(2x + 1 + 2x) + 2(x² - 1) = 0

<=> (4x + 1) + 2x² - 2 = 0

<=> 4x + 1 + 2x² - 2 = 0

<=> 2x² + 4x - 2 + 1 = 0

<=> 2x² + 4x - 1 = 0

<=> 2x² + 4x = 1

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

Vì 1 chỉ có tích là 1 . 1 nên:

<=> \(\left[{}\begin{matrix}2x=1\\x+2=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-1\end{matrix}\right.\)

\(a,\Leftrightarrow4x^2+4x+1-4x^2+2x^2-2=0\\ \Leftrightarrow2x^2+4x-1=0\\ \Leftrightarrow2\left(x^2+2x+1\right)-3=0\\ \Leftrightarrow2\left(x+1\right)^2-3=0\\ \Leftrightarrow\left(x+1\right)^2=\dfrac{3}{2}\\ \Leftrightarrow\left[{}\begin{matrix}x+1=\sqrt{\dfrac{3}{2}}\\x+1=-\sqrt{\dfrac{3}{2}}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-2-\sqrt{6}}{2}\\x=\dfrac{-2+\sqrt{6}}{2}\end{matrix}\right.\)

\(b,\left(x-2\right)\left(x+2\right)-\left(x+3\right)^2-2x-5=0\\ \Leftrightarrow x^2-4-x^2-6x-9-2x-5=0\\ \Leftrightarrow-8x=18\\ \Leftrightarrow x=-\dfrac{9}{4}\)

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

\(\Rightarrow\left(x-\sqrt{2}\right)\cdot\left(x+3\sqrt{2}-1\right)=0\Rightarrow x=\sqrt{2};x=1-3\sqrt{2}\)