Đa giác. Diện tích của đa giác

Câu 1:

Đặt \(x+1=a\). Khi đó \(x+3=a+2; x-1=a-2\).

PT đã cho tương đương với:

\((a+2)^4+(a-2)^4=626\)

\(\Leftrightarrow 2a^4+48a^2+32=626\)

\(\Leftrightarrow a^4+24a^2-297=0\)

\(\Leftrightarrow (a^2+12)^2=441\)

\(\Rightarrow a^2+12=\sqrt{441}=21\) (do \(a^2+12>0)\)

\(\Rightarrow a^2=9\Rightarrow a=\pm 3\)

Nếu $a=3$ thì \(x=a-1=2\)

Nếu $a=-3$ thì $x=a-1=-4$

Câu 2:

Đặt \(2x-1=a; x-1=b\). PT đã cho tương đương với:

\(a^3+b^3+(-a-b)^3=0\)

\(\Leftrightarrow a^3+b^3-(a+b)^3=0\)

\(\Leftrightarrow a^3+b^3-[a^3+b^3+3ab(a+b)]=0\)

\(\Leftrightarrow ab(a+b)=0\Rightarrow \left[\begin{matrix} a=0\\ b=0\\ a+b=0\end{matrix}\right.\)

\(\Rightarrow \left[\begin{matrix} 2x-1=0\\ x-1=0\\ 3x-2=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{1}{2}\\ x=1\\ x=\frac{2}{3}\end{matrix}\right.\)

Câu 3:

PT \(\Leftrightarrow x^3(2x-1)-6x^2(2x-1)+9x(2x-1)-2(2x-1)=0\)

\(\Leftrightarrow (2x-1)(x^3-6x^2+9x-2)=0\)

\(\Leftrightarrow (2x-1)[x^2(x-2)-4x(x-2)+(x-2)]=0\)

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

\(\Rightarrow \left[\begin{matrix} 2x-1=0\\ x-2=0\\ x^2-4x+1=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{1}{2}\\ x=2\\ (x-2)^2=3\end{matrix}\right.\)

\(\Rightarrow \left[\begin{matrix} x=\frac{1}{2}\\ x=2\\ x=2\pm \sqrt{3}\end{matrix}\right.\)

bạn ơi bạn ngi sai rùi kìa ko có F

\(a.\:\left(2x+3\right)^2-4\left(x-1\right)\left(x+1\right)=49\\ 4x^2+12x+9-4x^2+4=49\\ 12x=49-9\\ x=\dfrac{40}{12}=\dfrac{10}{3}\)

4x² +12x+9-4(x²-1)=49

4x²+12x+9-4x+4=49

12x+13=49

12x=36

x=3

\(\left(2x+3\right)^2-4\left(x-1\right)\left(x+1\right)=49\)

\(\Rightarrow\left(2x+3\right)^2-\left(4x-4\right)\left(x+1\right)=49\)

\(\Rightarrow4x^2+12x+9-4x^2-4x+4x+4\)

\(\Rightarrow\left(4x^2-4x^2\right)+\left(4x-4x\right)+\left(4+9\right)+12x=49\)

\(\Rightarrow12x+13=49\Rightarrow12x=36\Rightarrow x=3\)

\(c.\:\left(3x+4\right)^2-\left(3x+1\right)\left(3x-1\right)\\ =9x^2+24x+16-9x^2+1\\ 40x=-1\\ x=-\dfrac{1}{40}\)

\(d.\:\left(3x-1\right)^2-\left(3x-2\right)^2=0\\ \left(3x-1+3x-2\right)\left(3x-1-3x+2\right)=0\\ \left(6x-3\right)=0\\ x=\dfrac{1}{2}\)

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

c,\(\left(3x+4\right)^2-\left(3x-1\right)\left(3x+1\right)=49\)

\(\Rightarrow9x^2+24x+16-\left(9x^2-1\right)=49\)

\(\Rightarrow9x^2+24x+16-9x^2+1=49\)

\(\Rightarrow24x=49-1-16\)

\(\Rightarrow24x=32\Rightarrow x=\dfrac{4}{3}\)

d, \(\left(3x-1\right)^2-\left(3x-2\right)^2=0\)

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

\(\Rightarrow6x-3=0\Rightarrow6x=3\Rightarrow x=\dfrac{1}{2}\)

e, \(\left(2x+1\right)^2-\left(x-1\right)^2=0\)

\(\Rightarrow\left(2x+1-x+1\right)\left(2x+1+x-1\right)=0\)

\(\Rightarrow\left(x+2\right).3x=0\Rightarrow x.\left(x+2\right)=0\)

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

Chúc bạn học tốt!!!

a, Xem lại đề:

b, \(16x^2-\left(4x-5\right)^2=15\)

\(\Rightarrow16x^2-\left(16x^2-40x+25\right)=15\)

\(\Rightarrow16x^2-16x^2+40x-25=15\)

\(\Rightarrow40x=40\Rightarrow x=1\)

Chúc bạn học tốt!!!

\(a.\:\left(7x+3\right)^2-4\left(x-1\right)\left(x+1\right)=49\\ 49x^2+42x+9-4x^2+4=49\\ 45x^2+42x+13=49\\ x^2+\dfrac{42}{45}x+\dfrac{13}{45}=\dfrac{49}{45}\\ x^2+2.\dfrac{7}{15}x+\left(\dfrac{7}{15}\right)^2=\dfrac{49}{45}-\dfrac{13}{45}+\left(\dfrac{7}{15}\right)^2\\ \left(x+\dfrac{7}{15}\right)^2=\dfrac{229}{225}\\ \Rightarrow\left[{}\begin{matrix}x+\dfrac{7}{15}=\dfrac{229}{225}\\x+\dfrac{7}{15}=-\dfrac{229}{225}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{124}{225}\\x=-\dfrac{334}{225}\end{matrix}\right.\)

Bài 3:

\(x=-\dfrac{180218}{772265}\)

Let's go

Bài 1

đặt \(S_{ABO}=S_1\left(S_1>0\right);S_{CDO}=S_2\left(S_2>0\right);S_{BOC}=S_3\left(S_3>0\right);S_{ADO}=S_4\left(S_4>0\right)\)

Ta có

\(\left\{{}\begin{matrix}S_1+S_3=S_1+S_4=2,5\left(cm^2\right)\\S_2+S_3=S_2+S_4=5\left(cm^2\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2S_1+S_3+S_4=5\left(cm^2\right)\\2S_2+S_3+S_4=10\left(cm^2\right)\end{matrix}\right.\) \(\Rightarrow2\left(S_2-S_1\right)=5\left(cm^2\right)\) (*)

mặt khác ta chứng minh được \(\Delta AOB\text{ đồng dạng}\Delta COD\)

suy ra \(\dfrac{S_1}{S_2}=\left(\dfrac{AB}{CD}\right)^2=\dfrac{1}{4}\) (**)

(*);(**) \(\Rightarrow S_1=\dfrac{5}{6}\left(cm^2\right);S_2=\dfrac{10}{3}\left(cm^2\right);S_3=S_4=\dfrac{5}{3}\left(cm^2\right)\)

kẻ CF, DK vuông góc với AB

đặt AK=x(x>0) thì AF=4-x

ta có

\(\left(4-x\right)^2+2,5^2=6^2\)

\(\left[{}\begin{matrix}x=\dfrac{8+\sqrt{119}}{2}\left(n\right)\\x=\dfrac{8-\sqrt{119}}{2}\left(l\right)\end{matrix}\right.\)

vậy \(BD=\sqrt{\left(x+2\right)^2+2,5^2}\approx11,72cm\)

next

Bài 2: mình lười làm bài hình lắm nên mình tua qua luôn

finally

Bài 3: