Bài 2:

\(a,45+170+25+30\)

\(=\left(45+25\right)+\left(170+30\right)\)

\(=60+200=260\)

Bài 3:

\(a,\left(x-6\right).5=150\)

\(x-6=150:5\)

\(x-6=30\)

\(x=30+6\)

\(x=36\)

\(b,2^5.\left(3x-2\right)=2^3.2^6\)

\(2^5.\left(3x-2\right)=2^{3+6}\)

\(2^5.\left(3x-2\right)=2^9\)

\(3x-2=2^9:2^5\)

\(3x-2=2^4=16\)

\(3x=16+2\)

\(3x=22\)

\(x=22:3\)

\(x\approx7,3\)

\(c,100-7.\left(x-5\right)=51\)

\(7.\left(x-5\right)=100-51\)

\(7.\left(x-5\right)=49\)

\(x-5=49:7\)

\(x-5=7\)

\(x=7+5\)

\(x=12\)

Phần d) bạn thiếu dữ liệu ạ.

Giúp với ạ 

Giúp với ạ 

\(3^{2x}=\left(3^x\right)^2=5^2=25\) -> D

`a)[2x-3]/4-[4x-5]/3=[5-x]/6`

`<=>3(2x-3)-4(4x-5)=2(5-x)`

`<=>6x-9-16x+20=10-2x`

`<=>8x=1`

`<=>x=1/8`

(`b->c` mk đã làm r).

\(a,\Rightarrow4x^2-20x-4x^2+3x+4x-3=5\\ \Rightarrow-13x=8\Rightarrow x=-\dfrac{8}{13}\\ b,\Rightarrow3x^2-10x+8-3x^2+27x=-3\\ \Rightarrow17x=-11\Rightarrow x=-\dfrac{11}{17}\\ c,\Rightarrow\left(x+3\right)\left(2-x\right)=0\Rightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\\ d,\Rightarrow2x\left(4x^2-25\right)=0\\ \Rightarrow2x\left(2x-5\right)\left(2x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{5}\\x=-\dfrac{2}{5}\end{matrix}\right.\\ e,Sửa:\left(4x-3\right)^2-3x\left(3-4x\right)=0\\ \Rightarrow\left(4x-3\right)^2+3x\left(4x-3\right)=0\\ \Rightarrow\left(4x-3\right)\left(7x-3\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{7}\end{matrix}\right.\)

a.

4x(x-5) - (x-1)(4x-3)-5=0

 4x^2-20x-4x^2+3x+4x+3=0

(4x^2-4x^2)+(-20x+3x+4x)+3=0

 13x+3 = 0

13x=-3

x=-3/13

b,

(3x-4)(x-2)-3x(x-9)+3=0

3x^2-6x-4x+8 - 3x^2+27x+3=0

(3x^2-3x^2)+(-6x-4x+27x)+(8+3)=0

17x+11=0

17x=-11

x=-11/17

c, 2(x+3)-x^2-3x=0

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

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

TH1: x+3 = 0; x=-3

TH2: 2-x=0;x=2

\(a,\) \(5x\left(4-x\right)+\left(5x^2-12\right)=x+6\)

\(< =>20x-5x^2+5x^2-12-x-6=0\)

\(< =>19x-18=0\)

\(< =>x=\dfrac{18}{19}\)

\(b,\left(2x-7\right)\left(5+4x\right)-8\left(x^2-4x+5\right)=-30\)

\(< =>10x+8x^2-35-28x-8x^2+24x-40+30=0\)

\(< =>6x-45=0< =>x=\dfrac{45}{6}=7,5\)

a) \(5x\left(4-x\right)+\left(5x^2-12\right)=x+\Rightarrow6\\ \Leftrightarrow20x-5x^2+5x^2-12=x+6\\ \Leftrightarrow20x-12=x+6\\\Rightarrow20x-x=6+12\\ \Rightarrow19x=18\\ \Rightarrow x=\dfrac{18}{19}\)

b) \(\left(2x-7\right)\left(5+4x\right)-8\left(x^2-3x+5\right)=-30\\ \Rightarrow10x+8x^2-35-28x-8x^2+24x-40=-30\\ \Rightarrow6x-75=-30\\ \Rightarrow6x=45\\ \Rightarrow x=\dfrac{15}{2}\)

a)x=18/19

b)x=15/2

a) Ta có: \(2\left(3x+1\right)-4\left(5-2x\right)>2\left(4x-3\right)-6\)

\(\Leftrightarrow6x+2-20+8x>8x-6-6\)

\(\Leftrightarrow14x-18-8x+12>0\)

\(\Leftrightarrow6x-6>0\)

\(\Leftrightarrow6x>6\)

hay x>1

Vậy: S={x|x>1}

b) Ta có: \(9x^2-3\left(10x-1\right)< \left(3x-5\right)^2-21\)

\(\Leftrightarrow9x^2-30x+3< 9x^2-30x+25-21\)

\(\Leftrightarrow9x^2-30x+3-9x^2+30x-4< 0\)

\(\Leftrightarrow-1< 0\)(luôn đúng)

Vậy: S={x|\(x\in R\)}

Tất cả các bài này nếu lười suy nghĩ thì bình lên bậc 4 rồi dùng máy tính bỏ túi tìm nghiệm và phân tích nhân tử!

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

\(VT=\sqrt{3}\left[2\left(x^2-x+1\right)-\left(x^2+x+1\right)\right]\)

Có dạng đẳng cấp rồi.

2/ \(x^4+1=\left(x^2+1\right)^2-2x^2=\left(x^2-\sqrt{2}x+1\right)\left(x^2+\sqrt{2}x+1\right)\)

\(VT=\left(x^2+\sqrt{2}x+1\right)+3\left(x^2-\sqrt{2}x+1\right)\)-> dạng đẳng cấp

3/ tương tự: \(x^3+3x^2+4x+2=\left(x^2+2x+2\right)\left(x+1\right)\)

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

4/ Chuyển vế căn ở giữa, bình phương thu gọn rồi làm giống như 3 bài ở trên.

5/ Có lẽ tương tự

gõ latex đi bn!

a. 2^x + 70 = 74

<=> 2^x = 4

<=> x = 2

b. 120 - \(\dfrac{4x}{2}\) = 80

<=> 120 - 2x = 80

<=> 120 - 80 = 2x

<=> 2x = 40

<=> x = 20

c. (3x + 5)² = 400

<=> \(|3x+5|=\sqrt{400}\)

<=> \(|3x+5|=20\)

<=> \(\left[{}\begin{matrix}3x+5=20\\3x+5=-20\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{-25}{3}\end{matrix}\right.\)