Trong thùng còn số ki-lô-gam đường là:

\(28,75-10,5-8=10,25\) (ki-lô-gam đường)

a) Xét \(\Delta ABD\left(\widehat{ADB}=90^o\right)\) và \(\Delta ACE\left(\widehat{AEC}=90^o\right)\) có:

\(AB=AC\) (giả thiết)

\(\widehat{A}\) chung

\(\Rightarrow\Delta ABD=\Delta ACE\left(ch.gn\right)\)

\(\Rightarrow AD=AE\) (\(2\) cạnh tương ứng)

\(\Rightarrow\Delta ADE\) cân tại \(A\)

b) Vì \(\Delta ADE\) cân tại \(A\) (chứng minh trên)

\(\Rightarrow\widehat{AED}=\dfrac{180^o-\widehat{A}}{2}\) (1)

Vì \(\Delta ABC\) cân tại \(A\) (giả thiết)

\(\Rightarrow\widehat{ABC}=\dfrac{180^o-\widehat{A}}{2}\) (2)

Từ (1) và (2) \(\Rightarrow\widehat{AED}=\widehat{ABC}\) 

Mà \(2\) góc này ở vị trí đồng vị

\(\Rightarrow DE//BC\)

c) Ta có: \(\left\{{}\begin{matrix}\widehat{EBI}+\widehat{EIB}=90^o\\\widehat{DCI}+\widehat{DIC}=90^o\end{matrix}\right.\) (\(2\) góc phụ nhau)

Mà \(\widehat{EIB}=\widehat{DIC}\) (\(2\) góc đối đỉnh)

\(\Rightarrow\widehat{EBI}=\widehat{DCI}\)

Lại có: \(\left\{{}\begin{matrix}AE+EB=AB\left(E\in AB\right)\\AD+DC=AC\left(D\in AC\right)\end{matrix}\right.\)

Mà \(\left\{{}\begin{matrix}AB=AC\left(gt\right)\\AE=AD\left(cmt\right)\end{matrix}\right.\)

\(\Rightarrow EB=DC\)

Xét \(\Delta BEI\left(\widehat{BEI}=90^o\right)\) và \(\Delta CDI\left(\widehat{CDI}=90^o\right)\) có:

\(EB=DC\left(cmt\right)\)

\(\widehat{EIB}=\widehat{DIC}\) (\(2\) góc đối đỉnh)

\(\Rightarrow\Delta BEI=\Delta CDI\left(cgv.gn\right)\)

\(\Rightarrow BI=CI\) (\(2\) cạnh tương ứng)

Tàu ở độ sâu so với mực nước biển là:

\(155+58-70=143\left(m\right)\)

Ta có: \(A=\dfrac{4}{n-3}\left(n\in Z\right)\)

a) Để \(A\) là phân số thì \(n-3\ne0\Leftrightarrow n\ne3\)

b) Để \(A\in Z\Rightarrow\left(n-3\right)\inƯ\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)

\(\Rightarrow n\in\left\{4;3;5;1;7;-1\right\}\)

Vậy \(n\in\left\{4;3;5;1;7;-1\right\}\) thì \(A\in Z\)

a) \(\left(3x-2\right)\left(4x+5\right)=0\)

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

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

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

b) \(\left(2,3x-6,9\right)\left(0,1x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2,3x-6,9=0\\0,1x+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-20\end{matrix}\right.\)

c) \(\left(4x+2\right)\left(x^2+1\right)=0\)

Vì \(x^2+1\ge1>0\forall x\)

\(\Rightarrow4x+2=0\)

\(\Leftrightarrow x=-\dfrac{1}{2}\)

Vậy: \(S=\left\{-\dfrac{1}{2}\right\}\)

d) \(\left(2x+7\right)\left(x-5\right)\left(5x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+7=0\\x-5=0\\5x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=5\\x=-\dfrac{1}{5}\end{matrix}\right.\)

Vậy: \(S=\left\{-\dfrac{7}{2};5;-\dfrac{1}{5}\right\}\)

e) \(\left(x-1\right)\left(2x+7\right)\left(x^2+2\right)=0\)

Vì \(x^2+2\ge2>0\forall x\)

\(\Rightarrow\left(x-1\right)\left(2x+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+7=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{7}{2}\end{matrix}\right.\)

f) \(\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right)\)

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

\(\Leftrightarrow\left[\left(3x+2\right)\left(x+1\right)\right].\left(x-1-3x+2\right)=0\)

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

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

\(\Leftrightarrow\left[3x\left(x+1\right)+2\left(x+1\right)\right]\left(-2x+1\right)=0\)

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

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

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

Vậy: \(S=\left\{-1;-\dfrac{2}{3};\dfrac{1}{2}\right\}\)

a) \(\dfrac{-x}{4}=\dfrac{-9}{x}\)

\(\Rightarrow-x^2=-36\)

\(\Rightarrow x^2=36\)

\(\Rightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)

Vậy: \(x\in\left\{6;-6\right\}\)

b) \(\dfrac{5}{9}+\dfrac{x}{-1}=-\dfrac{1}{3}\)

\(\Rightarrow\dfrac{5}{9}+\dfrac{-9x}{9}=\dfrac{-3}{9}\)

\(\Rightarrow5-9x=-3\)

\(\Rightarrow-9x=-8\)

\(\Rightarrow x=\dfrac{8}{9}\)

Vậy: \(x=\dfrac{8}{9}\)

c) \(x:3\dfrac{1}{5}=1\dfrac{1}{2}\)

\(\Rightarrow x:\dfrac{16}{5}=\dfrac{3}{2}\)

\(\Rightarrow x=\dfrac{3}{2}.\dfrac{16}{5}\)

\(\Rightarrow x=\dfrac{24}{5}\)

Vậy: \(x=\dfrac{24}{5}\)

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

\(\Rightarrow\left(3x-1\right)^2=25\)

\(\Rightarrow\left[{}\begin{matrix}3x-1=5\\3x-1=-5\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}3x=6\\3x=-4\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{4}{3}\end{matrix}\right.\)

Vậy: \(x\in\left\{2;-\dfrac{4}{3}\right\}\)

Diện tích hình bình hành \(AFEB\) là:

\(45\times138=6210\left(cm^2\right)\)

Diện tích hình chữ nhật \(BEDC\) là:

\(138\times42=5796\left(cm^2\right)\)

Diện tích hình \(K\) là:

\(6210+5796=12006\left(cm^2\right)\)

\(BPT\Leftrightarrow\dfrac{2\sqrt{a}}{3}\le\dfrac{3}{3}\Leftrightarrow2\sqrt{a}\le3\Leftrightarrow a\le\dfrac{9}{4}\)

Ta có: \(A=-\dfrac{4}{n-1}\)

a) Để \(A\) là phân số thì \(n-1\ne0\Leftrightarrow n\ne1\)

b) Để \(A\in Z\) thì \(n-1\inƯ\left(-4\right)=\left\{1;-1;2;-2;4;-4\right\}\)

\(\Rightarrow n\in\left\{2;0;3;-1;5;-3\right\}\)

Bạn xem lại câu \(d\).