a. 3(2,2−0,3x)=2,6+(0,1x−4)3(2,2−0,3x)=2,6+(0,1x−4)

⇔6,6−0,9x=2,6+0,1x−4⇔6,6−2,6+4=0,1x+0,9x⇔x=8⇔6,6−0,9x=2,6+0,1x−4⇔6,6−2,6+4=0,1x+0,9x⇔x=8

Phương trình có nghiệm x = 8.

b. 3,6−0,5(2x+1)=x−0,25(2−4x)3,6−0,5(2x+1)=x−0,25(2−4x)

⇔3,6−x−0,5=x−0,5+x⇔3,6−0,5+0,5=x+x+x⇔3,6=3x⇔x=1,2⇔3,6−x−0,5=x−0,5+x⇔3,6−0,5+0,5=x+x+x⇔3,6=3x⇔x=1,2

Phương trình có nghiệm x = 1,2

a) 1,2-(x-0,8)= -2(0,9+x)

VT=-(x-2)

VP=\(-\frac{10x+9}{5}\)

pt trở thành:-(x-2)=\(-\frac{10x+9}{5}\)

<=>2-x=-2x-1,8

<=>5x=-19

<=>x=-3,8

a) 1,2-(x-0,8)= -2(0,9+x)

VT=-(x-2)

VP=$-\frac{10x+9}{5}$−10x+95 

pt trở thành:-(x-2)=$-\frac{10x+9}{5}$−10x+95 

<=>2-x=-2x-1,8

<=>5x=-19

<=>x=-3,8

b) 2,3x-2(0,7+2x)=3,6-1,7x

x=-2.25179981368525*10^16

c) 3(2,2-0,3x)=2,6+(0,1x-4)

Áp dụng tc tỉ lệ thức ta có:

\(\frac{66-9x}{10}=\frac{x-14}{10}\Rightarrow\left(66-9x\right)10=10\left(x-14\right)\)

VT=-30(3x-22)

pt trờ thành -30(3x-22)=10(x-14)

<=>660-90x=10x-140

<=>-100x=-800

<=>x=8

d) 3,6-0,5(2x+1)=x-0,25(2-4x)

Áp dụng tc tỉ lệ thức ta típ tục biến đổi được

\(\frac{31-10x}{10}=\frac{4x-1}{2}\Rightarrow\left(31-10x\right)2=10\left(4x-1\right)\)

VT=-2(10x-31)

pt trở thành -2(10x-31)=40x-10

<=>-60x=-72

<=>x=1,2

a)\(1,2-x+0,8=-1,8-2x\)

\(2-x=-1,8-2x\)

\(2x-x=-1,8-2\)

\(x=-3,8\)

Vậy S={-3,8}

b)\(2,3x-1,4-4x=3,6-1,7x\)

\(2,3x-4x+1,7x=3,6+1,4\)

0=5(vô lí)

Vậy S={\(\varnothing\)}

c)\(6,6-0.9=2,6+0,1x-4\)

\(5,7=0,1x-1,4\)

\(-4,3=0,1x\)

\(x=-43\)

Câu c đáng lẽ là như thế này chứ.

c, 3(2.2-0.3x)=2.6+(0.1x-4)

<=> 6.6-0.9x=2.6+0.1x-4

<=> 6.6-0.9x=0.1x-1.4

<=>-0.9x -0.1x =-8

<=> -x=-8

<=> x=8

Mình trả lời câu d luôn nhé.

d, 3.6-0.5(2x+1)=x-0.25(2-4x)

<=> 3.6-x-0.5=x-0.5+x

<=> 3.1-x=2x-0.5

<=>-x-2x=-3.6

<=> -3x=-3.6

<=> x= 1.2

\(a.ĐK:x\ne3;1\)

\(\Rightarrow\dfrac{1}{2\left(x-3\right)}+\dfrac{3x-10}{\left(x-1\right)\left(x-3\right)}=\dfrac{7}{2}\)

\(\Leftrightarrow\dfrac{\left(x-1\right)+2\left(3x-10\right)}{2\left(x-1\right)\left(x-3\right)}=\dfrac{7\left(x-1\right)\left(x-3\right)}{2\left(x-1\right)\left(x-3\right)}\)

\(\Leftrightarrow x-1+2\left(3x-10\right)=7\left(x-1\right)\left(x-3\right)\)

\(\Leftrightarrow x-1+6x-20=7\left(x^2-4x+3\right)\)

\(\Leftrightarrow7x-21=7x^2-28x+21\)

\(\Leftrightarrow7x^2-35x+42=0\)

\(\Leftrightarrow7\left(x^2-5x+6\right)=0\)

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

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

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

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

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

b.\(ĐK:x\ne2;4\)

\(\Rightarrow\dfrac{x-1}{x-2}-\dfrac{x+3}{4-x}=\dfrac{2}{\left(x-2\right)\left(4-x\right)}\)

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

\(\Leftrightarrow\left(x-1\right)\left(4-x\right)-\left(x+3\right)\left(x-2\right)=2\)

\(\Leftrightarrow4x-x^2-4+x-x^2+2x-3x+6-2=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=2\left(ktm\right)\end{matrix}\right.\)

a: \(\Leftrightarrow\dfrac{1}{2\left(x-3\right)}+\dfrac{3x-10}{\left(x-1\right)\left(x-3\right)}=\dfrac{7}{2}\)

\(\Leftrightarrow x-1+2\left(3x-10\right)=7\left(x-1\right)\left(x-3\right)\)

\(\Leftrightarrow7\left(x^2-4x+3\right)=x-1+6x-20=7x-21\)

\(\Leftrightarrow\left(x-3\right)\left(7x-7\right)-7\left(x-3\right)=0\)

=>(x-3)(7x-14)=0

=>x=3(loại) hoặc x=2(nhận)

b: \(\Leftrightarrow\left(x-1\right)\left(x-4\right)+\left(x+3\right)\left(x-2\right)=-2\)

\(\Leftrightarrow x^2-5x+4+x^2+x-6=-2\)

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

=>2x(x-2)=0

=>x=0(nhận) hoặc x=2(loại)

a. TH1:

\(\left\{{}\begin{matrix}x^2+3x-4< 0\\3-2x>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x< 1\\x>-4\end{matrix}\right.\\x>\dfrac{3}{2}\end{matrix}\right.\)

TH2:

\(\left\{{}\begin{matrix}x^2+3x-4>0\\3-2x< 0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x>1\\x< -4\end{matrix}\right.\\x< \dfrac{3}{2}\end{matrix}\right.\)

Vậy nghiệm của BPT:

\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x< 1\\x>-4\end{matrix}\right.\\x>\dfrac{3}{2}\end{matrix}\right.\)      \(\left\{{}\begin{matrix}\left[{}\begin{matrix}x>1\\x< -4\end{matrix}\right.\\x< \dfrac{3}{2}\end{matrix}\right.\)

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

⇔ \(\left[{}\begin{matrix}\left\{{}\begin{matrix}\left(2x-5\right)\left(x+2\right)< 0\\4x-3>0\end{matrix}\right.\\\left\{{}\begin{matrix}\left(2x-5\right)\left(x+2\right)>0\\4x-3< 0\end{matrix}\right.\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}\left\{{}\begin{matrix}-2< x< \dfrac{5}{2}\\x>\dfrac{3}{4}\end{matrix}\right.\\\left\{{}\begin{matrix}\left[{}\begin{matrix}x< -2\\x>\dfrac{5}{2}\end{matrix}\right.\\x< \dfrac{3}{4}\end{matrix}\right.\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}\dfrac{3}{4}< x< \dfrac{5}{2}\\x< -2\end{matrix}\right.\)

Vậy tập nghiệm của bất phương trình là

S = \(\left(\dfrac{3}{4};\dfrac{5}{2}\right)\cup\left(-\infty;-2\right)\)

b, Pt

⇔ \(\left\{{}\begin{matrix}x^2-5x+6=x^2+6x+5\\x\in R\backslash\left\{-1;2\right\}\end{matrix}\right.\)

⇔ x = \(\dfrac{1}{11}\)

Vậy S = \(\left\{\dfrac{1}{11}\right\}\)

a: =>3,6-1,7x=2,3-1,4-4=0,9-4=-3,1

=>1,7x=6,7

hay x=67/17

b: \(\Leftrightarrow30\left(5x+4\right)-15\left(3x+5\right)=24\left(4x+9\right)-40\left(x-9\right)\)

=>150x+120-45x-75=96x+216-40x+360

=>105x+45=56x+576

=>49x=531

hay x=531/49

tách nhỏ câu hỏi ra nhé dài quá