a)\(3:\left(2x+1\right)=72-69\)

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

\(2x+1=3:3\)

\(2x+1=1\)

\(2x=1-1\)

\(2x=0\)

\(x=0:2\)

\(x=0\)

các bài còn lại giống như câu a nha nếu ko biết thì comment lại minhf sẽ giải cho . Nhớ k cho mình nha

(35:x+3).19=165-13=152

35:x+3=152:19=8

35:x=8-3=5

x=35:5=7

2x+69.2=69.4

2x+138=276

2x = 276-138

2x = 138

x = 138:2

x = 69

2x-12-x = 0

<=> 2x-x-12 =0

(2x-x)-12=0

=> x-12=0

x = 0+12

x  =12

(x-7)(2x-8)=0

\(\Rightarrow\hept{\begin{cases}x-7=0\\2x-8=0\end{cases}}\Rightarrow\hept{\begin{cases}x=7\\x=4\end{cases}}\)

Vậy ...

a, 2X+69.2=69.4

2X+138=276

2X=276-138=138

X=138:2=69

b,2X-12-x=0 

2X-X=12-0

x=12

mik chi biet vay thoi con cau cuoi mik chiu

2x+69.2=69.4

2x=69.4-69.2

2x=69.(4-2)

2x=69.2

x=69.2/2

x=69

2x-12-x=0

2x-x=0+12

x=12

(x-7).(2x-8)=0

ta có hai trường hợp:  

trường hợp 1:     x-7=0

                          x=7

trường hợp 2:

2x-8=0

2x=8

x=4

bạn cố gắng hiểu nha. mình làm hơi tắt.

a)

⇔ x2 + 6x + 9 – 10 ≥ x2 + 3x + 2x + 6 – 4

⇔ x2 – x2 + 6x – 3x – 2x ≥ –9 + 10 + 6 – 4

⇔ x ≥ 3

Tập nghiệm: S = {x | x ≥ 3}.

Biểu diễn trên trục số:

Vậy tập nghiệm của bất phương trình là 

a) \(x^2-x+x=4\)

\(x^2=4\)

\(x=\pm2\)

b) \(3x\left(x-5\right)-2\left(x-5\right)=0\)

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

\(\left[{}\begin{matrix}x=5\\x=\dfrac{2}{3}\end{matrix}\right.\)

c) Ta có: \(a+b+c=5-3-2=0\)

\(\left[{}\begin{matrix}x=1\\x=\dfrac{c}{a}=\dfrac{-2}{5}\end{matrix}\right.\)

d) Đặt \(x^2=t\left(t\ge0\right)\) . Lúc đó phương trình trở thành :

\(t^2-11t+18=0\)

\(\left[{}\begin{matrix}t=9\left(tmđk\right)\\t=2\left(tmđk\right)\end{matrix}\right.\)

\(t=9\rightarrow x^2=9\rightarrow x=\pm3\)

\(t=2\rightarrow x^2=2\rightarrow x=\pm\sqrt{2}\)

a) 2x-12=18                   x=3

2x - 12 = 18................................

2x = 30 ........................................

x = 15.....................................

2x - 12 = 18................................

2x = 30 ........................................

x = 15.....................................

a:Ta có: \(x\left(x-1\right)+x=4\)

\(\Leftrightarrow x^2-x+x=4\)

\(\Leftrightarrow x^2=4\)

hay \(x\in\left\{2;-2\right\}\)

b: Ta có: \(3x\left(x-5\right)-2x+10=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{2}{3}\end{matrix}\right.\)

c: Ta có: \(5x^2-3x-2=0\)

\(\Leftrightarrow5x^2-5x+2x-2=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{2}{5}\end{matrix}\right.\)

d: Ta có: \(x^4-11x^2+18=0\)

\(\Leftrightarrow x^4-9x^2-2x^2+18=0\)

\(\Leftrightarrow x^2\left(x^2-9\right)-2\left(x^2-9\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\\x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\)

a) x(x-1)+x=4

⇔x²=4⇔\(x=\pm2\)

b)3x(x-5)-2x+10=0

⇔3x(x-5)-2(x-5)=0

⇔(x-5)(3x-1)=0

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{1}{3}\end{matrix}\right.\)

c)5x²-3x-2=0

⇔ 5x(x-1)+2(x-1)=0

⇔ (x-1)(5x+2)=0

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{2}{5}\end{matrix}\right.\)

d)x⁴-11x²+18=0

⇔ x²(x²-2)-9(x²-2)=0

⇔ (x²-2)(x²-9)=0

\(\Leftrightarrow\left[{}\begin{matrix}x^2=2\\x^2=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\pm\sqrt{2}\\x=\pm3\end{matrix}\right.\)

a: \(\left(2x-3\right)^2=\left|3-2x\right|\)

=>\(\left\{{}\begin{matrix}\left|2x-3\right|>=0\\\left(2x-3\right)^2=\left(2x-3\right)\end{matrix}\right.\Leftrightarrow\left(2x-3\right)^2-\left(2x-3\right)=0\)

=>\(\left(2x-3\right)\left(2x-3-1\right)=0\)

=>\(\left(2x-3\right)\left(2x-4\right)=0\)

=>\(\left[{}\begin{matrix}2x-3=0\\2x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=2\end{matrix}\right.\)

b: \(\left(x-1\right)^2+\left(2x-1\right)^2=0\)

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

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

\(\Delta=\left(-6\right)^2-4\cdot5\cdot2=36-20\cdot2=-4< 0\)

=>Phương trình vô nghiệm

c: ĐKXĐ: x>=0

\(x-2\sqrt{x}=0\)

=>\(\sqrt{x}\cdot\sqrt{x}-2\cdot\sqrt{x}=0\)

=>\(\sqrt{x}\left(\sqrt{x}-2\right)=0\)

=>\(\left[{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=4\left(nhận\right)\end{matrix}\right.\)

d: \(\left(x-1\right)^2+\dfrac{1}{7}=0\)

mà \(\left(x-1\right)^2+\dfrac{1}{7}>=\dfrac{1}{7}>0\forall x\)

nên \(x\in\varnothing\)

\(a,\Rightarrow3x^2-3x+6-2x-3x^2=0\\ \Rightarrow-5x=-6\Rightarrow x=\dfrac{6}{5}\\ b,\Rightarrow\left(x-1\right)\left(x-1+x+2\right)=0\\ \Rightarrow\left(x-2\right)\left(2x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{1}{2}\end{matrix}\right.\\ c,\Rightarrow x^2\left(2x+3\right)+\left(2x+3\right)=0\\ \Rightarrow\left(x^2+1\right)\left(2x+3\right)=0\Rightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\2x+3=0\end{matrix}\right.\\ \Rightarrow x=-\dfrac{3}{2}\\ d,\Rightarrow2x^2+x-6=0\\ \Rightarrow2x^2+4x-3x-6=0\\ \Rightarrow2x\left(x+2\right)-3\left(x+2\right)=0\\ \Rightarrow\left(2x-3\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-2\end{matrix}\right.\)

\(a,\Rightarrow\left(2x-1\right)\left(2x+1\right)-x\left(2x+1\right)=0\\ \Rightarrow\left(2x+1\right)\left(2x-1-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=1\end{matrix}\right.\\ b,\Rightarrow\left(x-3\right)\left(x-4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\\ c,\Rightarrow\left(x^2-8x+16\right)-10=0\\ \Rightarrow\left(x-4\right)^2-10=0\\ \Rightarrow\left(x-4-\sqrt{10}\right)\left(x-4+\sqrt{10}\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=4+\sqrt{10}\\x=4-\sqrt{10}\end{matrix}\right.\)