Ta có : (P) \(y=x^2\)

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

Xét phương trình hoành độ giao điểm của (P) và (d):

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

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

Vậy hoành độ giao điểm của (P) và (d) là 1 và \(\dfrac{-3}{2}\)

tìm tọa độ giao điểm?

phương trình hoành độ giao điểm:

x^2 = (-1/2) x + 3/2

<=> x = 1 hoặc x = -3/2

x = 1 -> y = 1

x = -3/2 => y = 9/4

=> tọa độ giao điểm A(1;1) ; B(-3/2;9/4)

a, \(\left(-2xy^2z^3\right)^3.\left(\dfrac{5}{2}xy^3\right)^2.\left(\dfrac{-4}{125}xy\right)\)

\(=\left(-2\right)^3.x^3.\left(y^2\right)^3.z^3.\left(\dfrac{5}{2}\right)^2.x^2.\left(y^3\right)^2.\dfrac{-4}{125}.x.y\)

\(=\left(-2\right)^3.\left(\dfrac{5}{2}\right)^2.\dfrac{-4}{125}.\left(x^3.x^2.x\right).\left(y^6.y^6.y\right).z^3\)

\(=\left(-8\right).\dfrac{25}{4}.\dfrac{-4}{125}.x^6.y^{13}.z^3\)

\(=1,6.x^6.y^{13}.z^3\)

a, \(\left(-2xy^2z^3\right).\left(\dfrac{5}{2}xy^3\right)^2.\left(\dfrac{-4}{125}xy\right)\)

= \(\left(-5x^2y^5z^3\right)^5.\left(\dfrac{-4}{125}xy\right)\)

= \(\left(\dfrac{4}{25}x^3y^6z^3\right)^5\)

b, \(2\dfrac{1}{3}x^2y^5-3\dfrac{2}{5}x^3y-1\dfrac{1}{2}x^2y^5+2\dfrac{2}{3}x^3y\)

= \(\dfrac{7}{3}x^2y^5-\dfrac{17}{5}x^3y-\dfrac{3}{2}x^2y^5+\dfrac{8}{3}x^3y\)

= \(\dfrac{5}{6}x^2y^5-\dfrac{11}{15}x^3y\)

b) Để \(\frac{2}{3\left|x-1\right|+4}\)đạt GTLN

=> 3|x - 1|+ 4 đạt giá trị nhỏ nhất

mà  3|x - 1| \(\ge0\forall x\)

=> 3|x - 1| + 4 \(\ge\)4

Dấu "=" xảy ra <=> x - 1 = 0

=> x = 1

Vậy GTLN của \(B=\frac{1}{2}\Leftrightarrow x=1\)

câu A đâu ?

Ý a) của bạn mình nghĩ là \(A=-\left|\frac{1}{2}x-3\right|+4\)

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

Ta có : \(\left|\frac{1}{2}x-3\right|\ge0\forall x\Rightarrow-\left|\frac{1}{2}x-3\right|\le0\forall x\)

Cộng 4 vào mỗi vế

=> \(A=-\left|\frac{1}{2}x-3\right|+4\le4\forall x\)

Dấu " = " xảy ra <=> \(\frac{1}{2}x-3=0\)=> x = 6

Vậy A_Max = 4 khi x = 6 

a) Áp dụng tính chất dãy tỉ số bằng nhau ta được:

X/3 = y/4 = x/3 + y/4 = 28/7 = 4

=> x = 4 × 3 = 12

=> y = 4 × 4 = 16

Vậy x = 12, y = 16

B) Áp dụng tính chất dãy tỉ số bằng nhau ta được:

X/2 = y/(-5) = x/2 - y/(-5) = (-7)/7 = -1

=> x = -1 × 2 = -2

=> y = -1 × -5 = 5

Vậy x = -2, y = 5

C) làm tương tự như bài a, b

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(1.x-\dfrac{2}{3}\times\left(x+9\right)=1\)

\(x-\dfrac{2}{3}\times x-6=1\)

\(x\times\left(1-\dfrac{2}{3}\right)=7\)

\(x\times\dfrac{1}{3}=7\)

\(x=21\)

\(2.x-\dfrac{11}{15}=\dfrac{3+x}{5}\)

\(\dfrac{15x}{15}-\dfrac{11}{15}=\dfrac{9+3x}{15}\)

\(15x-11=9+3x\)

\(12x=20\)

\(x=\dfrac{5}{3}\)

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

               \(\dfrac{4}{5}\).\(x\) = \(\dfrac{3}{5}\)+ \(\dfrac{2}{5}\)

                 \(\dfrac{4}{5}\).\(x\) = 1

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

b, - \(\dfrac{3}{7}\) - \(\dfrac{4}{7}\): \(x\) = \(\dfrac{2}{5}\)

              \(\dfrac{4}{7}\): \(x\) = - \(\dfrac{3}{7}\) - \(\dfrac{2}{5}\)

                \(\dfrac{4}{7}\): \(x\) = - \(\dfrac{29}{35}\)

                  \(x\) = \(\dfrac{4}{7}\): (- \(\dfrac{29}{35}\) )

                  \(x\) = - \(\dfrac{20}{29}\)

c, \(\dfrac{4}{7}\).\(x\) + \(\dfrac{2}{3}\) = - \(\dfrac{1}{5}\)

     \(\dfrac{4}{7}\).\(x\)         = -\(\dfrac{1}{5}\) - \(\dfrac{2}{3}\)

      \(\dfrac{4}{7}\).\(x\)       = - \(\dfrac{13}{15}\)

           \(x\)     = - \(\dfrac{13}{15}\): \(\dfrac{4}{7}\)

            \(x\)    = - \(\dfrac{91}{60}\)

d, \(\dfrac{5}{7}\): \(x\) - 1 = \(\dfrac{2}{3}\)

     \(\dfrac{5}{7}\): \(x\)     = \(\dfrac{2}{3}\)+ 1

       \(\dfrac{5}{7}\): \(x\)   = \(\dfrac{5}{3}\)

              \(x\)  = \(\dfrac{5}{7}\): \(\dfrac{5}{3}\)

               \(x\) = \(\dfrac{3}{7}\)

\(a,\dfrac{2}{3}x-\dfrac{2}{5}=\dfrac{1}{2}x-\dfrac{1}{3}\\ \Rightarrow\dfrac{2}{3}x-\dfrac{1}{2}x-\dfrac{2}{5}=-\dfrac{1}{3}\\ \Rightarrow x\left(\dfrac{2}{3}-\dfrac{1}{2}\right)-\dfrac{2}{5}=-\dfrac{1}{3}\\ \Rightarrow x\dfrac{1}{6}=-\dfrac{11}{15}\\ \Rightarrow x=-\dfrac{22}{5}\\ b,\dfrac{1}{3}x+\dfrac{2}{5}.\left(x+1\right)=0\\ \Rightarrow\dfrac{1}{3}x+\left(x+1\right)=-\dfrac{2}{5}\\ \Rightarrow\dfrac{1}{3}x=-\dfrac{2}{5}-\left(x+1\right)\\ \Rightarrow\dfrac{1}{3}x=-\dfrac{7}{5}-x\\ \Rightarrow\dfrac{1}{3}.2x=-\dfrac{7}{5}\\ \Rightarrow2x=-\dfrac{21}{5}\\ \Rightarrow x=-\dfrac{21}{10}.\)

a,     \(\dfrac{3}{7}\)\(x\)- \(\dfrac{2}{3}\)\(x\)    = \(\dfrac{10}{21}\)

    (\(\dfrac{3}{7}\) - \(\dfrac{2}{3}\)) \(\times\) \(x\)  =  \(\dfrac{10}{21}\)

     - \(\dfrac{5}{21}\) \(\times\) \(x\)      = \(\dfrac{10}{21}\)

                 \(x\)      = \(\dfrac{10}{21}\) : (-\(\dfrac{5}{21}\))

                 \(x\)      = -2 

b, \(\dfrac{7}{35}\) : (\(x-\dfrac{1}{3}\)) = - \(\dfrac{2}{25}\)

            \(x\) - \(\dfrac{1}{3}\)    =  \(\dfrac{7}{35}\) : (- \(\dfrac{2}{25}\))

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

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

              \(x\)      = - \(\dfrac{13}{6}\)

c, 3.(\(x\) - \(\dfrac{1}{2}\)) - 5.(\(x\) + \(\dfrac{3}{5}\)) = - \(x\)+ \(\dfrac{1}{5}\)

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

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

              \(x\)           = - \(\dfrac{47}{10}\)

\(a,\dfrac{3}{7}x-\dfrac{2}{3}x=\dfrac{10}{21}\\ \Rightarrow x\left(\dfrac{3}{7}-\dfrac{2}{3}\right)=\dfrac{10}{21}\\ \Rightarrow x.-\dfrac{5}{21}=\dfrac{10}{21}\\ \Rightarrow x=-2\\ b,\dfrac{7}{35}:\left(x-\dfrac{1}{3}\right)=-\dfrac{2}{25}\\ \Rightarrow\dfrac{1}{5}:\left(x-\dfrac{1}{3}\right)=-\dfrac{2}{25}\\ \Rightarrow x-\dfrac{1}{3}=-\dfrac{5}{2}\\ \Rightarrow x=-\dfrac{13}{6}\\ c,3.\left(x-\dfrac{1}{2}\right)-5.\left(x+\dfrac{3}{5}\right)=-x+\dfrac{1}{5}\\ \Rightarrow3x-\dfrac{3}{2}-5x+5=-x+\dfrac{1}{5}\)

\(\Rightarrow x\left(3-5\right)-\dfrac{3}{2}+5=-x+\dfrac{1}{5}\\ \Rightarrow-2x-\dfrac{13}{2}=-x+\dfrac{1}{5}\\ \Rightarrow-x-\dfrac{13}{5}=\dfrac{1}{5}\\ \Rightarrow x=\dfrac{1}{5}-\dfrac{13}{5}\\ \Rightarrow x=-\dfrac{12}{5}.\)

a,73​x−32​x=2110​⇒x(73​−32​)=2110​⇒x.−215​=2110​⇒x=−2b,357​:(x−31​)=−252​⇒51​:(x−31​)=−252​⇒x−31​=−25​⇒x=−613​c,3.(x−21​)−5.(x+53​)=−x+51​⇒3x−23​−5x+5=−x+51​

⇒�(3−5)−32+5=−�+15⇒−2�−132=−�+15⇒−�−135=15⇒�=15−135⇒�=−125.⇒x(3−5)−23​+5=−x+51​⇒−2x−213​=−x+51​⇒−x−513​=51​⇒x=51​−513​⇒x=−512​.