\((4.x-15)^3=5^3=4.x-15=5=4.x=20=>x=5\)

Bài 1 bạn tự là nhé dễ lắm :))

Bài 2 :

Ta có :

\(4x-15=-75-x\)
\(\Leftrightarrow\)\(4x+x=-75+15\)

\(\Leftrightarrow\)\(5x=-60\)

\(\Leftrightarrow\)\(x=\frac{-60}{5}\)

\(\Leftrightarrow\)\(x=-12\)

Bài 3 :

Ta có :

\(12⋮\left(x-3\right)\)\(\Rightarrow\)\(\left(x-3\right)\inƯ\left(12\right)\)

Mà \(Ư\left(12\right)=\left\{1;-1;2;-2;3;-3;4;-4;6;-6;12;-12\right\}\)

Suy ra : ( lập bảng )

Vậy \(x\in\left\{4;2;5;1;6;0;7;-1;9;-3;15;-9\right\}\)

Chúc bạn học tốt 

\(x^3+4x^2+4x+3\)

\(=\left(x^3+3x^2\right)+\left(x^2+3x\right)+\left(x+3\right)\)

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

Phân tích đa thức thành nhân tử à bạn?

thank bạn <3

a) (x-3).(y+5) = 11 = 1.11 = (-1).(-11)

TH1: x - 3 = 1 => x = 4

y + 5 = 11 => y = 6

TH2: x - 3 = 11 => x = 14

y+5=1 => y = -4

...

bn tự lm típ nhé!

b) |x-1| +|3+y| = 0

=> |x-1| = 0 =>x-1 = 0 => x = 1

|3+y| = 0 => 3+y = 0=> y = - 3

c) ta có: 4x+3 chia hết cho x - 1

=> 4x -4+7 chia hế cho x - 1

4.(x-1) + 7 chia hết cho x - 1

mà 4.(x-1) chia hết cho x - 1

=> 7 chia hết cho x - 1

=> x - 1 thuộc Ư(7)={1;-1;7;-7}

...

rùi bn lập bảng xét giá trị hộ mk nha!!
 

Cái này chắc là đề THCS chứ không phải toán lớp 2 bạn à. !! >_<

1) Ta có: \(4x^2-1=\left(2x+1\right).\left(3x-5\right)\)

\(\Leftrightarrow\left(2x+1\right).\left(2x-1\right)-\left(2x+1\right).\left(3x-5\right)=0\)

\(\Leftrightarrow\left(2x+1\right).\left[\left(2x-1\right)-\left(3x-5\right)\right]=0\)

\(\Leftrightarrow\left(2x+1\right).\left(2x-1-3x+5\right)=0\)

\(\Leftrightarrow\left(2x+1\right).\left(4-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=0\\4-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-1\\-x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\frac{1}{2}\left(TM\right)\\x=4\left(TM\right)\end{matrix}\right.\)

Vậy \(x=-\frac{1}{2}\) hoặc \(x=4\)

2) Ta có: \(\left(x+1\right)^2=4.\left(x^2-2x+1\right)\)

\(\Leftrightarrow\left(x+1\right)^2-\left[2.\left(x-1\right)\right]^2=0\)

\(\Leftrightarrow\left[\left(x+1\right)+2.\left(x-1\right)\right].\left[\left(x+1\right)-2.\left(x-1\right)\right]=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\3-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=1\\-x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{3}\left(TM\right)\\x=3\left(TM\right)\end{matrix}\right.\)

Vậy \(x=\frac{1}{3}\) hoặc \(x=3\)

3) Ta có: \(2x^3+5x^2-3x=0\)

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

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

\(\Leftrightarrow x.\left[x.\left(2x-1\right)+3.\left(2x-1\right)\right]=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\2x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(TM\right)\\x=-3\left(TM\right)\\x=-\frac{1}{2}\left(TM\right)\end{matrix}\right.\)

Vậy \(x=0\) hoặc \(x=-3\) hoặc \(x=-\frac{1}{2}\)

4) Ta có: \(2x=3x-2\)

\(\Leftrightarrow2x-3x=-2\)

\(\Leftrightarrow-x=-2\)

\(\Leftrightarrow x=2\left(TM\right)\)

Vậy \(x=2\)

5) Ta có: \(x+15=3x-1\)

\(\Leftrightarrow x-3x=-1-15\)

\(\Leftrightarrow-2x=-16\)

\(\Leftrightarrow x=8\left(TM\right)\)

Vậy \(x=8\)

6) Ta có: \(2-x=0,5x-4\)

\(\Leftrightarrow-x-0,5x=-4-2\)

\(\Leftrightarrow-1,5x=-6\)

\(\Leftrightarrow x=4\left(TM\right)\)

Vậy \(x=4\)

1) 4x²-1=(2x+1)(3x-5)

<=> (2x-1)(2x+1)-(2x+1)(3x-5)=0

<=> (2x+1)(2x-1-3x+5)=0

<=> (2x+1)(4-x)=0

<=>\([^{2x+1=0}_{4-x=0}< =>[^{2x=-1}_{x=4}< =>[^{x=\frac{-1}{2}}_{x=4}\)

2) (x+1)²= 4(x²-2x+1)

<=> x²+2x+1-4(x²-2x+1)=0

<=> x²+2x+1-4x²+8x-4=0

<=> -3x²+10x-3=0

<=> -3x²+x+9x-3=0

<=> -x(3x-1)+3(3x-1)=0

<=> (3x-1)(3-x)=0

<=> \([^{3x-1=0}_{3-x=0}< =>[^{3x=1}_{x=3}< =>[^{x=\frac{1}{3}}_{x=3}\)

3) 2x³+5x²-3x=0

<=> 2x(x²+\(\frac{5}{2}x-\frac{3}{2})=0\)

<=> 2x\(\left[x^2+2.\frac{5}{4}x+\frac{25}{16}-\left(\frac{25}{16}+\frac{3}{2}\right)\right]=0\)

<=> 2x\(\left[\left(x+\frac{5}{4}\right)^2-\frac{49}{16}\right]=0\)

<=> 2x\(\left(x+\frac{5}{4}-\frac{7}{4}\right)\left(x+\frac{5}{4}+\frac{7}{4}\right)=0\)

<=> x\(\left(x-\frac{1}{2}\right)\left(x+3\right)=0\)

<=>\(\left[{}\begin{matrix}x=0\\x=\frac{1}{2}\\x=-3\end{matrix}\right.\)

4) 2x=3x-2

<=> 2x-3x=-2

<=> -x=-2

<=> x=2

5) x+15=3x-1

<=> x-3x=1-15

<=> -2x=-14

<=> x=-14:-2

<=> x=7

6) 2-x=0,5x-4

<=> -x-0,5x=-4-2

<=> -1,5x=-6

<=> x= -6: -1,5

<=> x=4

học tốt nghen

1, \(4x^2-1=\left(2x+1\right)\left(3x-5\right)\)

\(\Leftrightarrow4x^2-1=6x^2-10x+3x-5\)

\(\Leftrightarrow4x^2-1=6x^2-7x-5\)

\(\Leftrightarrow6x^2-7x-5-4x^2+1=0\)

\(\Leftrightarrow2x^2-7x-4=0\)

\(\Leftrightarrow\left(2x^2-8x\right)-\left(x-4\right)=0\)

\(\Leftrightarrow2x\left(x-4\right)-\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(2x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-4=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\frac{1}{2}\end{matrix}\right.\)

Vậy phương trình có tập nghiệm là \(S=\left\{4;\frac{1}{2}\right\}\)

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

\(\Leftrightarrow x^2+2x+1=4x^2-8x+4\\ \Leftrightarrow4x^2-8x+4-x^2-2x-1=0\\ \Leftrightarrow3x^2-10x+3=0\\ \Leftrightarrow\left(3x^2-x\right)-\left(9x-3\right)=0\\ \Leftrightarrow x\left(3x-1\right)-3\left(3x-1\right)=0\\ \Leftrightarrow\left(3x-1\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x-1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\x=3\end{matrix}\right.\)

Vậy phương trình có tập nghiệm là \(S=\left\{3;\frac{1}{3}\right\}\)

3,

\(2x^3+5x^2-3x=0\\ \Leftrightarrow x\left(2x^2+5x-3\right)=0\\ \Leftrightarrow x\left[\left(2x^2+x\right)-\left(6x+3\right)\right]=0\\ \Leftrightarrow x\left[x\left(2x+1\right)-3\left(2x+1\right)\right]=0\\ \Leftrightarrow x\left(2x+1\right)\left(x-3\right)=0\\ \left[{}\begin{matrix}x=0\\2x+1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\frac{1}{2}\\x=3\end{matrix}\right.\)

Vậy phương trình có tập nghiệm là \(S=\left\{0;-\frac{1}{2};3\right\}\)

3^x-2-19=62

=> 3^x-2=62+19

=> 3^x-2=81

=> 3^x-2=3⁴

=> x-2=4

=> x=4+2

=> x=6

10.3^x=3¹¹⁷+3¹¹⁵

=> 10.3^x=3¹¹⁵.(3²+1)

=> 10.3^x=3¹¹⁵.(9+1)

=> 3^x.10=3¹¹⁵.10

=> 3^x=3¹¹⁵

=> x=115

2448-(5x+48)=2000

=> 5x+48=2448-2000

=> 5x+48=448

=> 5x=448-48

=> 5x=400

=> x=400:5

=> x=80

1)\(3^{x-2}=62+19=81=3^4\)

=> x-  2 = 4 

=> x = 6

2) \(2448-\left(5x+48\right)=2000\)

  5x + 48   = 448

  5x          = 448 - 48

 5x            = 400

 x              = 80 

3) \(10.3^x=3^{115}\left(3^2+1\right)=3^{115}.10\)  

=> x =115 

Bà con ơi thắng Trần câu nào cùng được 2,3 lik-e cả chắc trao đổi lik-e 

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

Có: (2x+1)² ≥ 0 ∀x => (2x+1)² + 1 ≥ 1 > 0 (đpcm)

3. -x² + 4x - 5 = -(x² - 4x + 4) - 1 = -(x - 2)^2 - 1

Có: -(x-2)^2 ≤ 0 => -(x-2)^2 -1 ≤ - 1 < 0 (đpcm)

7. (x+2)(x-5) + 15 = x² - 3x + 5 = (x² - 2.x.\(\dfrac{3}{2}\)+ \(\dfrac{9}{4}\)) + \(\dfrac{11}{4}\)

= ( x - \(\dfrac{3}{2}\))^2 + \(\dfrac{11}{4}\) \(\ge\dfrac{11}{4}>0\left(đpcm\right)\)