Bài 1: 

a) Ta có: \(\dfrac{17}{6}-x\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)

\(\Leftrightarrow\dfrac{17}{6}-x^2+\dfrac{7}{6}x-\dfrac{7}{4}=0\)

\(\Leftrightarrow-x^2+\dfrac{7}{6}x+\dfrac{13}{12}=0\)

\(\Leftrightarrow-12x^2+14x+13=0\)

\(\Delta=14^2-4\cdot\left(-12\right)\cdot13=196+624=820\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{14-2\sqrt{205}}{-24}=\dfrac{-7+\sqrt{205}}{12}\\x_2=\dfrac{14+2\sqrt{2015}}{-24}=\dfrac{-7-\sqrt{205}}{12}\end{matrix}\right.\)

b) Ta có: \(\dfrac{3}{35}-\left(\dfrac{3}{5}-x\right)=\dfrac{2}{7}\)

\(\Leftrightarrow\dfrac{3}{5}-x=\dfrac{3}{35}-\dfrac{10}{35}=\dfrac{-7}{35}=\dfrac{-1}{5}\)

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

ai giúp mik vs

Bài 3: 

a) Ta có: \(2x-3=x+\dfrac{1}{2}\)

\(\Leftrightarrow2x-x=\dfrac{1}{2}+3\)

\(\Leftrightarrow x=\dfrac{7}{2}\)

b) Ta có: \(4x-\left(x+\dfrac{1}{2}\right)=2x-\left(\dfrac{1}{2}-5\right)\)

\(\Leftrightarrow3x-\dfrac{1}{2}-2x+\dfrac{1}{2}-5=0\)

\(\Leftrightarrow x=5\)

X = \(\dfrac{4}{7}:\dfrac{3}{5}=\dfrac{20}{21}\)

Học tốt

X = 

\(1)x+\frac{5}{6}\times2\frac{2}{5}-1\frac{1}{4}=35\%\)

 \(x+\frac{5}{6}\times\frac{12}{5}-\frac{5}{4}=\frac{7}{12}\)

\(x+\frac{5}{6}\times\frac{12}{5}=\frac{7}{12}+\frac{5}{4}\)

\(x+\frac{5}{6}.\frac{12}{5}=\frac{8}{5}\)

\(x+\frac{5}{6}=\frac{8}{5}:\frac{12}{5}\)

\(x+\frac{5}{6}=\frac{2}{3}\)

\(x=\frac{2}{3}-\frac{5}{6}\)

\(x=-\frac{1}{6}\)

HỌC TỐT !

\(2\)) \(\left|x-\frac{1}{2}\right|-\frac{3}{4}=0\)

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

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

           \(x-\frac{1}{2}\)          \(=\frac{3}{4}\)hoặc \(-\frac{3}{4}\)

Ta xét 2 trường hợp :

  Trường hợp 1 :         \(x-\frac{1}{2}=\frac{3}{4}\)

                                    \(x\)         \(=\frac{3}{4}+\frac{1}{2}\)

                                    \(x\)           \(=\frac{5}{4}\)

Trường hợp 2 :           \(x-\frac{1}{2}=-\frac{3}{4}\)

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

                                   \(x\)             \(=-\frac{1}{4}\)

Vậy \(x\in\text{{}\frac{5}{4};-\frac{1}{4}\)}

có giải chi tiết ko bạn

54 + 68,7 x 99 + 14,7

= 68,7 + 68,7 x 99

= 68,7 x 100

= 6870

54 + 68,7 x 99 + 14,7

= 68,7 + 68,7 x 99

= 68,7 x 100

= 6870

54 + 68,7 x 99 + 14,7

= 68,7 + 68,7 x 99

= 68,7 x 100

= 6870

=>2/3:x=-7-1/3=-22/3

=>x=-2/3:22/3=-2/3*3/22=-1/11

\(\dfrac{1}{3}\)+\(\dfrac{2}{3}\):x=-7

     \(\dfrac{2}{3}\):x=(-7)-\(\dfrac{1}{3}\)

     \(\dfrac{2}{3}\):x=\(\dfrac{-22}{3}\)

         x=\(\dfrac{2}{3}\):\(\dfrac{-22}{3}\)

         x=\(\dfrac{-1}{11}\) 

vậy x=\(\dfrac{-1}{11}\)

\(\dfrac{1}{3}+\dfrac{2}{3}:x=-7\\ \Rightarrow\dfrac{2}{3}:x=-7-\dfrac{1}{3}\\ \Rightarrow\dfrac{2}{3}:x=-\dfrac{21}{3}-\dfrac{1}{3}\\ \Rightarrow\dfrac{2}{3}:x=\dfrac{-22}{3}\\ \Rightarrow x=\dfrac{2}{3}:-\dfrac{22}{3}\\ \Rightarrow x=\dfrac{2}{3}\times-\dfrac{3}{22}\\ \Rightarrow x=-\dfrac{1}{11}\)

Đề có cho đa thức P(x) không bạn?

Vì P(x) chia cho đa thức bậc 2 nên dư là đa thức bậc 1

Gọi đa thức ấy là \(ax+b\)

\(\Leftrightarrow P\left(x\right)=\left(x^2-4x+3\right)\cdot a\left(x\right)+ax+b\\ \Leftrightarrow P\left(x\right)=\left(x-1\right)\left(x-3\right)\cdot a\left(x\right)+ax+b\)

\(P\left(1\right)=3\Leftrightarrow a+b=3\\ P\left(3\right)=7\Leftrightarrow3a+b=7\)

Từ đó ta có hệ \(\left\{{}\begin{matrix}a+b=3\\3a+b=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2a=4\\a+b=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=1\end{matrix}\right.\)

Vậy đa thức dư là \(2x+1\)

x : 7/11 =  5/7

x = 5/7 x 7/11

x = 5/11

x=5/11

x= 5/11