\(2\left(x-3\right)+3x+0,5=\dfrac{3}{4}\\ \Leftrightarrow2x-6+3x+\dfrac{1}{2}=\dfrac{3}{4}\\ \Leftrightarrow x\left(2+3\right)=\dfrac{3}{4}-\dfrac{1}{2}+6\\ \Leftrightarrow5x=\dfrac{25}{4}\\ \Leftrightarrow x=\dfrac{25}{4}:5=\dfrac{5}{4}\\ ---\\ 4^{x+2}+4^x=272\\ \Leftrightarrow4^x\left(4^2+1\right)=272\\ \Leftrightarrow4^x.17=272\\ \Leftrightarrow4^x=\dfrac{272}{17}=16=4^2\\ Vậy:x=2\\ ----\\ \left(1,2-5x\right)\left(2\dfrac{1}{8}+\dfrac{1}{2}x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}1,2-5x=0\\2,125+0,5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=1,2\\0,5x=-2,125\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1,2}{5}=0,24\\x=\dfrac{-2,125}{0,5}=-4,25\end{matrix}\right.\)

a) \(2\left(x-3\right)+3x+0,5=\dfrac{3}{4}\)

\(\Rightarrow2x-6+3x+\dfrac{1}{2}=\dfrac{3}{4}\)

\(\Rightarrow5x-6=\dfrac{3}{4}-\dfrac{1}{2}\)

\(\Rightarrow5x-6=\dfrac{1}{4}\)

\(\Rightarrow5x=\dfrac{1}{4}+6\)

\(\Rightarrow5x=\dfrac{25}{4}\)

\(\Rightarrow x=\dfrac{25}{4}:5\)

\(\Rightarrow x=\dfrac{5}{4}\)

b) \(4^{x+2}+4^x=272\)

\(\Rightarrow4^x\cdot4^2+4^x\cdot1=272\)

\(\Rightarrow4^x\cdot\left(16+1\right)=272\)

\(\Rightarrow4^x\cdot17=272\)

\(\Rightarrow4^x=16\)

\(\Rightarrow4^x=4^2\)

\(\Rightarrow x=2\)

c) \(\left(1,2-5x\right)\left(2\dfrac{1}{8}+\dfrac{1}{2}x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}1,2-5x=0\\\dfrac{15}{8}+\dfrac{1}{2}x=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}5x=1,2\\\dfrac{1}{2}x=-\dfrac{15}{8}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1,2}{5}\\x=-\dfrac{15}{8}:\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{6}{25}\\x=-\dfrac{15}{4}\end{matrix}\right.\)

a,3x-0,75=-1,2

3x=-1,2+-0,75

3x=-9/20

x=-9/20:3

x=-3/20

kl:....

b,2/3/5:(2x-1/2)=1,3

2/3/5??????

c,3/4x-5/6=7/12x-5/4

3/4x-7/12x=-5/4+5/6

9/12x-7/12x=-15/12+10/12

2/12x=-5/12

2:12x=-5/12

12x=2:-5/12

12x=-24/5

x=-24/5:12

x=-2/5

kl:....

chúc bn học tốt

mở dấu trị tuyệt đối ra rồi tính như bình thường

Chịu :)

S=n(n+1)mũ 2  trên   4

1.

\(x^4-6x^2-12x-8=0\)

\(\Leftrightarrow x^4-2x^2+1-4x^2-12x-9=0\)

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

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

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

\(\Leftrightarrow x=1\pm\sqrt{5}\)

3.

ĐK: \(x\ge-9\)

\(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)

\(\Leftrightarrow\left(x^2-x+1\right)\left(\sqrt{x+9}+x^2-9\right)=0\)

\(\Leftrightarrow\sqrt{x+9}+x^2-9=0\left(1\right)\)

Đặt \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{x+9}\\x=\sqrt{x+9}-1\end{matrix}\right.\)

\(\Leftrightarrow...\)

2.

ĐK: \(x\ne\dfrac{2\pm\sqrt{2}}{2};x\ne\dfrac{-2\pm\sqrt{2}}{2}\)

\(\dfrac{x}{2x^2+4x+1}+\dfrac{x}{2x^2-4x+1}=\dfrac{3}{5}\)

\(\Leftrightarrow\dfrac{1}{2x+\dfrac{1}{x}+4}+\dfrac{1}{2x+\dfrac{1}{x}-4}=\dfrac{3}{5}\)

Đặt \(2x+\dfrac{1}{x}+4=a;2x+\dfrac{1}{x}-4=b\left(a,b\ne0\right)\)

\(pt\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{3}{5}\left(1\right)\)

Lại có \(a-b=8\Rightarrow a=b+8\), khi đó:

\(\left(1\right)\Leftrightarrow\dfrac{1}{b+8}+\dfrac{1}{b}=\dfrac{3}{5}\)

\(\Leftrightarrow\dfrac{2b+8}{\left(b+8\right)b}=\dfrac{3}{5}\)

\(\Leftrightarrow10b+40=3\left(b+8\right)b\)

\(\Leftrightarrow\left[{}\begin{matrix}b=2\\b=-\dfrac{20}{3}\end{matrix}\right.\)

TH1: \(b=2\Leftrightarrow...\)

TH2: \(b=-\dfrac{20}{3}\Leftrightarrow...\)

\(a.1,2-\left(x-0,8\right)=-2\left(0,9+x\right)\\\Leftrightarrow1,2-x+0,8=-1,8-2x\\ \Leftrightarrow-x+2x=-1,2-0,8-1,8\\ \Leftrightarrow x=-3,8\)

Vậy nghiệm của phương trình trên là \(-3,8\)

\(b.2,3x-2\left(0,7+2x\right)=3,6-1,7x\\ \Leftrightarrow2,3x-1,4-4x=3,6-1,7x\\ \Leftrightarrow2,3x-4x+1,7x=1,4+3,6\\ \Leftrightarrow0x=5\)

\(\Rightarrow\)Vô nghiệm

\(c.5-\left(x-6\right)=4\left(3-2x\right)\\ \Leftrightarrow5-x+6=12-8x\\ \Leftrightarrow-x+8x=-5-6+12\\ \Leftrightarrow7x=1\\\Leftrightarrow x=\frac{1}{7}\)

Vậy nghiệm của phương trình trên là \(\frac{1}{7}\)

\(d.3,6-0,5\left(2x+1\right)=x-0,25\left(2-4x\right)\\ \Leftrightarrow3,6-x-0,5=x-0,5+x\\\Leftrightarrow -x-x-x=-3,6-0,5+0,5\\ \Leftrightarrow-3x=-3,6\\\Leftrightarrow x=1,2\)

Vậy nghiệm của phương trình trên là \(1,2\)

\(e.\left(x-3\right)\left(x+4\right)-2\left(3x-2\right)=\left(x-4\right)^2\\ \Leftrightarrow x^2+4x-3x-12-6x+4=x^2-8x+16\\\Leftrightarrow x^2-x^2+4x-3x-6x+8x=12-4+16\\ \Leftrightarrow3x=24\\ \Leftrightarrow x=8\)

Vậy nghiệm của phương trình trên là \(8\)

thứ nhất bn đăng sai môn 

thứ hai bn giải r đăng lmj :???

Thứ nhất đang sai môn 

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5gj5fg

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5j4fj

😱😱😱😱😱😱😱😱😱😱😱

\(\left|2+3x\right|=\left|4x-3\right|\)

\(\Leftrightarrow\orbr{\begin{cases}2+3x=4x-3\\2+3x=3-4x\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\\x=\frac{1}{7}\end{cases}}\)

Vậy \(x\in\left\{\frac{1}{7};5\right\}\)

\(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=1-4x\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}\)

Vậy \(x\in\left\{\frac{1}{11};\frac{3}{5}\right\}\)

\(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)

\(\Leftrightarrow\left|\frac{5}{4}x-\frac{7}{2}\right|=\left|\frac{5}{8}x+\frac{3}{5}\right|\)

Giải tiếp tương tự

Sau đó giải tiếp câu còn lại

a, \(\left|2+3x\right|=\left|4x-3\right|\)

\(\Rightarrow\orbr{\begin{cases}2+3x=4x-3\\2+3x=-4x+3\end{cases}\Rightarrow\orbr{\begin{cases}3x-4x=-3-2\\3x+4x=3-2\end{cases}\Rightarrow}\orbr{\begin{cases}-x=-5\\7x=1\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=\frac{1}{7}\end{cases}}}}\)

Câu b tương tự

c, \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)

\(\Rightarrow\left|\frac{5}{4}x-\frac{7}{2}\right|=\left|\frac{5}{8}x+\frac{3}{5}\right|\)

\(\Rightarrow\orbr{\begin{cases}\frac{5}{4}x-\frac{7}{2}=\frac{5}{8}x+\frac{3}{5}\\\frac{5}{4}x-\frac{7}{2}=-\frac{5}{8}x-\frac{3}{5}\end{cases}\Rightarrow\orbr{\begin{cases}\frac{5}{4}x-\frac{5}{8}x=\frac{3}{5}+\frac{7}{2}\\\frac{5}{4}x+\frac{5}{8}x=-\frac{3}{5}+\frac{7}{2}\end{cases}\Rightarrow}\orbr{\begin{cases}\frac{5}{8}x=\frac{41}{10}\\\frac{15}{8}x=\frac{29}{10}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}}\)

d, \(\left|\frac{7}{5}x+\frac{3}{2}\right|-\left|\frac{4}{3}-\frac{1}{4}\right|=0\)

\(\Rightarrow\left|\frac{7}{5}x+\frac{3}{2}\right|-\frac{13}{12}=0\)

\(\Rightarrow\left|\frac{7}{5}x+\frac{3}{2}\right|=\frac{13}{12}\)

Đến đây dễ rồi, tự làm tiếp :)

P/s: Ko chắc, sai ib với t :v