\(\frac{16}{8}+8\left(17-3x\right)=\frac{10^4}{10^3}\)

\(2+8\left(17-3x\right)=10\)

\(8\left(17-3x\right)=10-2\)

\(8\left(17-3x\right)=8\)

\(17-3x=\frac{8}{8}\)

\(17-3x=1\)

\(3x=17-1\)

\(3x=16\)

\(x=\frac{16}{3}\)

16 : 8 + 8.(17 - 3x) = 10⁴ : 10³

2 + (8.17 - 8.3x) = 10

=> 136 - 24x = 10 - 2

=> 136 - 24x = 8

=> 24x = 136 - 8

=> 24x = 128

=> \(x=\frac{128}{24}=\frac{16}{3}\)

Vậy \(x=\frac{16}{3}\)

16 : 8 + 8 ( 17 - 3x ) = 10⁴ : 10³

    2    + 136 - 16x = 10^4-3

        138 - 16x      = 10

                  16x      = 138 - 10

                   16x     =      128

                     x       = 128 : 16

                     x       = 8

Vậy x = 8

1. \(6x^3-8=40\\ 6x^3=48\\ x^3=8\\ \Rightarrow x=2\)Vậy x = 2

2. \(4x^5+15=47\\ 4x^5=32\\ x^5=8\\ \Rightarrow x\in\varnothing\left(\text{vì }x\in N\right)\)Vậy x ∈ ∅

3. \(2x^3-4=12\\ 2x^3=16\\ x^3=8\\ \Rightarrow x=2\)Vậy x = 2

4. \(5x^3-5=0\\ 5x^3=5\\ x^3=1\\ \Rightarrow x=1\)Vậy x = 1

5. \(\left(x-5\right)^{2016}=\left(x-5\right)^{2018}\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\x-5=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)Vậy \(x\in\left\{5;6\right\}\)

6. \(\left(3x-2\right)^{20}=\left(3x-1\right)^{20}\\ \Rightarrow3x-2=3x-1\\ 3x-3x=2-1\\ 0=1\left(\text{vô lí}\right)\)Vậy x ∈ ∅

7. \(\left(3x-1\right)^{10}=\left(3x-1\right)^{20}\\ \left(3x-1\right)^{10}=\left[\left(3x-1\right)^2\right]^{10}\\ \Rightarrow\left(3x-1\right)^2=3x-1\\ \left(3x-1\right)^2-\left(3x-1\right)=0\\ \left(3x-1\right)\left[\left(3x-1\right)-1\right]=0\\ \left(3x-1\right)\left(3x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x-1=0\\3x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x=1\\3x=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{1}{3}\left(\text{loại vì }x\in N\right)\\x=\frac{2}{3}\left(\text{loại vì }x\in N\right)\end{matrix}\right.\)Vậy x ∈ ∅

8. \(\left(2x-1\right)^{50}=2x-1\\ \left(2x-1\right)^{50}-\left(2x-1\right)=0\\ \left(2x-1\right)\left[\left(2x-1\right)^{49}-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}2x-1=0\\\left(2x-1\right)^{49}=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=1\\2x-1=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\frac{1}{2}\left(\text{loại vì }x\in N\right)\\x=1\left(t/m\right)\end{matrix}\right.\)Vậy x = 1

9. \(\left(\frac{x}{3}-5\right)^{2000}=\left(\frac{x}{3}-5\right)^{2008}\\ \left(\frac{x}{3}-5\right)^{2008}-\left(\frac{x}{3}-5\right)^{2000}=0\\ \left(\frac{x}{3}-5\right)^{2000}\left[\left(\frac{x}{3}-5\right)^8-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}\left(\frac{x}{3}-5\right)^{2000}=0\\\left(\frac{x}{3}-5\right)^8=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\frac{x}{3}-5=0\\\frac{x}{3}-5=1\\\frac{x}{3}-5=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\frac{x}{3}=5\\\frac{x}{3}=6\\\frac{x}{3}=4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\cdot3=15\\x=6\cdot3=18\\x=4\cdot3=12\end{matrix}\right.\)Vậy \(x\in\left\{15;18;12\right\}\)

\(1.6x^3-8=40\\ \Leftrightarrow6x^3=48\\ \Leftrightarrow x^3=8\Leftrightarrow x^3=2^3=\left(-2\right)^3\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

Vậy \(x\in\left\{2;-2\right\}\)

\(2.4x^3+15=47\) (T nghĩ đề là mũ 3)

\(\Leftrightarrow4x^3=32\Leftrightarrow x^3=8=2^3=\left(-2\right)^3\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

Vậy \(x\in\left\{2;-2\right\}\)

Câu 3, 4 tương tự nhé.

\(5.\left(x-5\right)^{2016}=\left(x-5\right)^{2018}\\ \Leftrightarrow\left(x-5\right)^{2018}-\left(x-5\right)^{2016}=0\\ \Leftrightarrow\left(x-5\right)^{2016}\left[\left(x-5\right)^2-1\right]=0\\ \Leftrightarrow\left(x-5\right)^{2016}\left(x-5-1\right)\left(x-5+1\right)=0\\ \Leftrightarrow\left(x-5\right)^{2016}\left(x-6\right)\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-5\right)^{2016}=0\\x-6=0\\x-4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x-5=0\\x=6\\x=4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=6\\x=4\end{matrix}\right.\)

Vậy \(x\in\left\{4;5;6\right\}\)

a) 27 . 3^x = 3¹⁰

27 . 3^x = 59049

3^x = 59049 : 27

3^x = 2187

3^x = 3⁷

x = 7

b) 60 + 3x = 9¹⁰ : 9⁸

60 + 3x = 9²

60 + 3x = 81

3x = 81 - 60

3x = 21

  x = 21 : 3

  x = 7

c) 23 - 3(4+x ) = 3

3( 4 + x ) = 23 - 3

3( 4 + x ) = 20

Đề bài sai nha bạn

a) \(27.3^x=3^{10}\)

\(3^3.3^x=3^{10}\)

\(3^{3+x}=3^{10}\)

\(\Rightarrow3+x=10\)

\(\Rightarrow x=7\)

vậy \(x=7\)

b) \(60+3x=9^{10}:9^8\)

\(60+3x=9^{10-8}\)

\(60+3x=9^2\)

\(60+3x=81\)

\(3x=81-60\)

\(3x=21\)

\(x=7\)

vậy \(x=7\)

c) \(23-3\left(4+x\right)=3\)

\(23-12-3x=3\)

\(11-3x=3\)

\(3x=11-3\)

\(3x=8\)

\(x=\frac{8}{3}\)

vậy \(x=\frac{8}{3}\)

a) 27 . 3^x = 3¹⁰

27 . 3^x = 59049

3^x = 59049 : 27

3^x = 2187

3^x = 3⁷

x = 7

b) 60 + 3x = 9¹⁰ : 9⁸

60 + 3x = 9²

60 + 3x = 81

3x = 81 - 60

3x = 21

  x = 21 : 3

  x = 7

c) 23 - 3(4+x ) = 3

3( 4 + x ) = 23 - 3

3( 4 + x ) = 20

Đề bài sai nha bạn

P/s tham khảo nha

Tìm x hả bạn ?

a ) \(\left(3x+\frac{1}{4}\right)^3=-27\)

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

\(\Rightarrow3x+\frac{1}{4}=-3\)

\(\Rightarrow3x=-3-\frac{1}{4}=-\frac{13}{4}\)

\(\Rightarrow x=-\frac{13}{4}:3=-\frac{13}{12}\)

Vậy x = \(-\frac{13}{12}\)

cho em hoi câu này xin các anh chị:

10mux x+4y = 2013

Cảm ơn Hoác Thiên Kình nhé .Thanh you 

(x-3)(x² + 3x + 9) - (3x-17) = x(x² - 8)

<=> x³ - 27 - 3x + 17 = x³ - 8x

<=> 5x - 10 = 0

<=> 5x = 10

<=> x = 2

Vậy ...

\(H=4x^2+4x+2=\left(2x+1\right)^2+1>0\)

\(K=4x^2+3x+2=4\left(x^2+2.\frac{3}{8}x+\frac{9}{64}\right)+\frac{23}{16}\)

      \(=4\left(x+\frac{3}{8}\right)^2+\frac{23}{16}>0\)

\(L=2x^2+3x+4=2\left(x^2+2.\frac{3}{4}x+\frac{9}{16}\right)+\frac{23}{8}\)

      \(=2\left(x+\frac{3}{4}\right)^2+\frac{23}{8}>0\)

Bạn viết rõ ra câu h đc ko giúp nha thanks bạn nhiều Đườnh Quỳnh Giang 

Phân tích đa thức thành nhân tử ?

Ta có: \(P=\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2\)

Đặt \(x^2+4x+8=y\)

Khi đó: 

\(P=y^2+3xy+2x^2\)

\(P=\left(y^2+xy\right)+\left(2xy+2x^2\right)\)

\(P=y\left(x+y\right)+2x\left(x+y\right)\)

\(P=\left(x+y\right)\left(2x+y\right)\)

\(P=\left(x^2+5x+8\right)\left(x^2+6x+8\right)\)

\(P=\left(x+2\right)\left(x+4\right)\left(x^2+5x+8\right)\)

\(4)D=x^2+x+1\)

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

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

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

Vậy biểu thức trên luôn nhận giá trị dương với mọi giá trị của x.

Các câu khác lm tương tự nhé.

Cho góp ý xíu: lần sau bn đưa từng câu một lên diễn đàn thì sẽ có câu trả lời nhanh hơn là đưa cùng một lúc như thế này đấy

hok tốt~

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

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

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

\(2\left(x+1\right)^2\ge0\forall x\Rightarrow2\left(x+1\right)^2+1\ge1>0\forall x\)( đpcm )

\(G=3x^2-5x+3=3\left(x^2-\frac{5}{3}x+\frac{25}{36}\right)+\frac{11}{12}=3\left(x-\frac{5}{6}\right)^2+\frac{11}{12}\)

\(3\left(x-\frac{5}{6}\right)^2\ge0\forall x\Rightarrow3\left(x-\frac{5}{6}\right)^2+\frac{11}{12}\ge\frac{11}{12}>0\forall x\)( đpcm )

\(H=4x^2+4x+2=4\left(x^2+x+\frac{1}{4}\right)+1=4\left(x+\frac{1}{2}\right)^2+1\)

\(4\left(x+\frac{1}{2}\right)^2\ge0\forall x\Rightarrow4\left(x+\frac{1}{2}\right)^2+1\ge1>0\forall x\)( đpcm )

\(K=4x^2+3x+2=4\left(x^2+\frac{3}{4}x+\frac{9}{64}\right)+\frac{23}{16}=4\left(x+\frac{3}{8}\right)^2+\frac{23}{16}\)

\(4\left(x+\frac{3}{8}\right)^2\ge0\forall x\Rightarrow4\left(x+\frac{3}{8}\right)^2+\frac{23}{16}\ge\frac{23}{16}>0\forall x\)( đpcm )

\(L=2x^2+3x+4=2\left(x^2+\frac{3}{2}x+\frac{9}{16}\right)+\frac{23}{8}=2\left(x+\frac{3}{4}\right)^2+\frac{23}{8}\)

\(2\left(x+\frac{3}{4}\right)^2\ge0\forall x\Rightarrow2\left(x+\frac{3}{4}\right)^2+\frac{23}{8}\ge\frac{23}{8}>0\forall x\)( đpcm )

1) x³ - 4x² - 8x + 8 

Thử với x = -2 ta có : (-2)³ - 4.(-2)² - 8.(-2) + 8 = 0

Vậy -2 là nghiệm của đa thức . Theo hệ quả của định lí Bézout thì đa thức trên chia hết cho x + 2

Thực hiện phép chia x³ - 4x² - 8x + 8 cho x + 2 ta được x² - 6x + 4

=> x³ - 4x² - 8x + 8 = ( x + 2 )( x² - 6x + 4 )

2) 3x² + 13x - 10

= 3x² + 15x - 2x - 10

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

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

3) x( 2x - 7 ) - 7 - 4x + 14 = 0

<=> 2x² - 7x - 4x + 7 = 0

<=> 2x² - 11x + 7 = 0

<=> 2( x² - 11/2x + 121/16 ) - 65/8 = 0

<=> 2( x - 11/4 )² = 65/8

<=> ( x - 11/4 )² = 65/16

<=> ( x - 11/4 )² = \(\left(\pm\sqrt{\frac{65}{16}}\right)^2=\left(\pm\frac{\sqrt{65}}{4}\right)^2\)

<=> \(\orbr{\begin{cases}x-\frac{11}{4}=\frac{\sqrt{65}}{4}\\x-\frac{11}{4}=\frac{-\sqrt{65}}{4}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{11+\sqrt{65}}{4}\\x=\frac{11-\sqrt{65}}{4}\end{cases}}\)

4) 2x³ + 3x² + 2x + 2 = 0 ( chịu không làm được ((: )