\(A=\left(x+\frac{4}{7}\right)^{24}+\frac{-12}{293}\)

Ta có \(\left(x+\frac{4}{7}\right)^{24}\ge0\forall x\Rightarrow\left(x+\frac{4}{7}\right)^{24}+\frac{-12}{293}\ge\frac{-12}{293}\)

Đẳng thức xảy ra <=> x + 4/7 = 0 => x = -4/7

=> MinA = -12/293 <=> x = -4/7

\(B=-\left(x+\frac{1}{6}\right)^{26}-\left(x+y+\frac{3}{8}\right)^{422}+5,98\)

Ta có \(\hept{\begin{cases}-\left(x+\frac{1}{6}\right)^{26}\le0\forall x\\-\left(x+y+\frac{3}{8}\right)^{442}\le0\forall x,y\end{cases}}\Rightarrow-\left(x+\frac{1}{6}\right)^{26}-\left(x+y+\frac{3}{8}\right)+5,98\le5,98\)

Đẳng thức xảy ra <=> \(\hept{\begin{cases}x+\frac{1}{6}=0\\x+y+\frac{3}{8}=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-\frac{1}{6}\\y=-\frac{5}{24}\end{cases}}\)

=> MaxB = 5, 98 <=> x = -1/6 ; y = -5/24

Vì AB // CD nên \(\hept{\begin{cases}\widehat{A}+\widehat{D}=180^0\\\widehat{B}+\widehat{C}=180^0\end{cases}}\)(định lí hình thang)

Mà \(\widehat{A}=5\widehat{D}\)=> \(\widehat{5D}+\widehat{D}=180^0\)=> \(6\widehat{D}=180^0\)=> \(\widehat{D}=30^0\)(1)

Thay (1) vào \(\widehat{A}=5\widehat{D}\)ta có :

\(\widehat{A}=5\cdot30^0=150^0\)

Lại có : \(\widehat{B}=4\widehat{C}\)

=> \(4\widehat{C}+\widehat{C}=180^0\)

=> \(5\widehat{C}=180^0\)

=> \(\widehat{C}=36^0\)(2)

Thay (2) vào \(\widehat{B}=4\widehat{C}\)ta có :

=> \(\widehat{B}=4\cdot36^0=144^0\)

Vậy : ^A = 150⁰ , ^B = 144⁰ , ^C = 36⁰ , ^D = 30⁰

Bài 1 :

\(a,\frac{15}{34}+\frac{7}{21}+\frac{19}{34}-1\frac{7}{21}+2\)

\(=\left(\frac{15}{34}+\frac{19}{34}\right)+\left(\frac{7}{21}-1\frac{7}{21}\right)+2\)

\(=1-1+2\)

\(=2\)

Học tốt

Bài 1 : 

a, 15/34+7/21+19/34-1 7/21+2

 = 15/34+1/3+19/34-4/3+2

 = (15/34+19/34)+(1/3-4/3)+2

 = 1+(-1)+2=0+2=2

a) 2( x - 1 )² + ( x + 3 )² = 3( x - 2 )( x + 1 )

<=> 2( x² - 2x + 1 ) + x² + 6x + 9 = 3( x² - x - 2 )

<=> 2x² - 4x + 2 + x² + 6x + 9 = 3x² - 3x - 6

<=> 2x² - 4x + x² + 6x - 3x² + 3x = -6 - 2 - 9

<=> 5x = -17

<=> x = -17/5

b) ( x - 1 )² - 2( x - 3 ) = ( x + 1 )²

<=> x² - 2x + 1 - 2x + 6 = x² + 2x + 1

<=> x² - 2x - 2x - x² - 2x = 1 - 1 - 6

<=> -6x = -6

<=> x = 1 

c) ( x - 3 )³ - ( x - 3 )( x² + 3x + 9 ) + 6( x + 1 )² + 3x² = -33

<=> x³ - 9x² + 27x - 27 - ( x³ - 3³ ) + 6( x² + 2x + 1 ) + 3x² = -33

<=> x³ - 9x² + 27x - 27 - x³ + 27 + 6x² + 12x + 6 + 3x² = -33

<=> x³ - 9x² + 27x - x³ + 6x² + 12x + 3x² = -33 - 27 + 27 - 6

<=> 39x = -39

<=> x = -1

a) Đặt \(a=x-1\)\(\Rightarrow\)\(\hept{\begin{cases}x+3=a+4\\x-2=a-1\\x+1=a+2\end{cases}}\)

    Ta có: \(2a^2+\left(a+4\right)^2=3.\left(a-1\right)\left(a+2\right)\)

        \(\Leftrightarrow2a^2+a^2+4a+4=3.\left(a^2+a-2\right)\)

        \(\Leftrightarrow3a^2+4a+4=3a^2+3a-6\)

        \(\Leftrightarrow a=-10\)

         \(\Rightarrow x-1=-10\)

        \(\Leftrightarrow x=-9\)

Vậy \(S=\left\{-9\right\}\)

b)  Đặt \(b=x-1\)\(\Rightarrow\)\(\hept{\begin{cases}x-3=b-2\\x+1=b+2\end{cases}}\)

     Ta có: \(b^2-2.\left(b-2\right)=\left(b+2\right)^2\)

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

         \(\Leftrightarrow-6b=0\) 

         \(\Leftrightarrow b=0\)

          \(\Rightarrow x-1=0\)

         \(\Leftrightarrow x=1\)

Vậy \(S=\left\{1\right\}\)

c) Ta có: \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2+3x^2=-33\)

        \(\Leftrightarrow\left(x-3\right)^3-\left(x-3\right)^3+6\left(x^2+2x+1\right)+3x^2+33=0\)

        \(\Leftrightarrow6x^2+12x+6+3x^2+33=0\)

       \(\Leftrightarrow9x^2+12x+39=0\)

       \(\Leftrightarrow\left(9x^2+12x+4\right)+35=0\)

       \(\Leftrightarrow\left(3x+2\right)^2+35=0\)

   Vì \(\left(3x+2\right)^2\ge0\forall x\)\(\Rightarrow\)\(\left(3x+2\right)^2+35\ge35>0\forall x\)

         mà \(\left(3x+2\right)^2+35=0\)

   \(\Rightarrow\)\(\left(3x+2\right)^2+35=0\)vô nghiệm

Vậy \(S=\varnothing\)

Bài 3 :

\(a,\left(\frac{2}{5}\right)^2+5\frac{1}{2}:\left(4,5-2\right)+\frac{2^3}{\left(-4\right)}\)

\(=\frac{4}{25}+\frac{11}{2}:\frac{5}{2}+\frac{8}{\left(-4\right)}\)

\(=\frac{4}{25}+\frac{11}{5}-2\)

\(=\frac{59}{25}-2\)

\(=\frac{9}{25}\)

b, Giống câu a

Học tốt

a) Nhân 2 vế của đẳng thức đầu cho 2³---> \(1^3.2^3+2^3.2^3+...+10^3.2^3=3025.2^3\)

                                                                  \(\Rightarrow2^3+4^3+...+20^3=3025.8=24200\)

b Chia 2 vế của đẳng thức đầu cho 2³---> \(\frac{1^3}{2^3}+\frac{2^3}{2^3}+...+\frac{10^3}{2^3}=\frac{3025}{2^3}\)

                                                               \(\Rightarrow0,5^3+1^3+2^3+...+5^3=\frac{3025}{8}=378,125\)

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

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

\(\Leftrightarrow-x^3+5x^2-5x=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}-x=0\\x^2-5x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\\left(x-\frac{5}{2}\right)^2-\frac{5}{4}=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\\left(x-\frac{5}{2}\right)^2=\frac{5}{4}\end{cases}\Leftrightarrow}\hept{\begin{cases}x=0\\x-\frac{5}{2}=\frac{\sqrt{5}}{2}\\x-\frac{5}{2}=-\frac{\sqrt{5}}{2}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=0\\x=\frac{5+\sqrt{5}}{2}\\x=\frac{5-\sqrt{5}}{2}\end{cases}}\)

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

\(\Leftrightarrow x\left(x-2-x^2+4x-3\right)=0\)

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

\(\Leftrightarrow x\left(x-\frac{5+\sqrt{5}}{2}\right)\left(x-\frac{5-\sqrt{5}}{2}\right)=0\)

=> \(x\in\left\{0;\frac{5+\sqrt{5}}{2};\frac{5-\sqrt{5}}{2}\right\}\)

b) \(\left(2x-5\right)\left(x+3\right)-\left(x-1\right)\left(2x+3\right)=0\)

\(\Leftrightarrow2x^2+x-15-2x^2-x+3=0\)

\(\Leftrightarrow-12=0\left(vn\right)\)

c) \(\left(x-2\right)\left(x^2+2x+8\right)-x^3-2x+1=0\)

\(\Leftrightarrow x^3+4x-16-x^3-2x+1=0\)

\(\Leftrightarrow2x=15\)

\(\Rightarrow x=\frac{15}{2}\)

Cài này EZ mà !!!