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

\(<=>2x(x-1)-3(x-1)=0\)

\(<=>(2x-3)(x-1)=0\)
th1 \(2x-3=0<=>x=3/2\)

th2 \(X-1=0<=>x=1\)

pt có tập nghiệm S={3/2;1}

\(2x^3+3x^2-8x+3=0\\ \Rightarrow\left(2x^3-2x^2\right)+\left(5x^2-5x\right)-\left(3x-3\right)=0\\ \Rightarrow2x^2\left(x-1\right)+5x\left(x-1\right)-3\left(x-1\right)=0\\ \Rightarrow\left(x-1\right)\left(2x^2+5x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-1=0\\2x^2+5x-3=0\end{matrix}\right.\)

\(x-1=0\\ \Rightarrow x=1\)

\(2x^2+5x-3=0\\ \Rightarrow\left(2x^2+6x\right)-\left(x+3\right)=0\\ \Rightarrow2x\left(x+3\right)-\left(x+3\right)=0\\ \Rightarrow\left(x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+3=0\\2x-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)

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

\(2x^3+3x-8x+3=0\)

\(\Leftrightarrow2x^3-5x+3=0\)

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

\(\Leftrightarrow2x\left(x-1\right)\left(x+1\right)-3\left(x-1\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\2x^2+2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{-1-\sqrt{7}}{2}\\x=\dfrac{-1+\sqrt{7}}{2}\end{matrix}\right.\)

3:

a: Xét ΔACH vuông tại H và ΔBCA vuông tại A có

góc C chung

=>ΔACH đồng dạng vơi ΔBCA

b: Xét ΔABC vuông tại A có AH là đường cao

nên AH^2=HB*HC

a: ĐKXĐ: x>=1

\(\sqrt{4x-4}+\sqrt{9x-9}=25\)

=>\(2\sqrt{x-1}+3\sqrt{x-1}=25\)

=>\(5\sqrt{x-1}=25\)

=>x-1=25

=>x=26(nhận)

b: \(\sqrt{4x^2-4x+1}=x-2\)

=>\(\sqrt{\left(2x-1\right)^2}=x-2\)

=>\(\left\{{}\begin{matrix}x>=2\\\left(2x-1\right)^2=\left(x-2\right)^2\end{matrix}\right.\)

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

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

c: \(\sqrt{x^2+4x+4}=2x-3\)

=>\(\sqrt{\left(x+2\right)^2}=2x-3\)

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

=>\(\left\{{}\begin{matrix}x>=\dfrac{3}{2}\\\left(x-5\right)\left(3x-1\right)=0\end{matrix}\right.\Leftrightarrow x=5\)

d: ĐKXĐ: 8-x>=0

=>x<=8

\(\sqrt{8-x}=x-2\)

=>\(\left\{{}\begin{matrix}x-2>=0\\\left(x-2\right)^2=8-x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2< =x< =8\\x^2-4x+4+x-8=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2< =x< =8\\x^2-3x-4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2< =x< =8\\\left(x-4\right)\left(x+1\right)=0\end{matrix}\right.\)

=>x=4

e: ĐKXĐ: x>=3

\(\sqrt{x-3}>2\)

=>x-3>4

=>x>7

f: ĐKXĐ: x>=0

\(\dfrac{4}{\sqrt{x}+1}>1\)

=>\(\dfrac{4}{\sqrt{x}+1}-1>0\)

=>\(\dfrac{4-\sqrt{x}-1}{\sqrt{x}+1}>0\)

=>\(3-\sqrt{x}>0\)

=>\(\sqrt{x}< 3\)

=>0<=x<9

a: CH=16^2/24=256/24=32/3(cm)

BC=24+32/3=104/3cm

AC=căn 32/3*104/3=16/3*căn 13(cm)

b: BC=12^2/6=144/6=24cm

CH=24-6=18cm

AC=căn 18*24=12*căn 3(cm)

1 trang

d)powerpoint

C1: có 3 trang tính

C2:c. Excel

C3: Thanh công thức dùng để nhập công thức hoặc nhập dữ liệu

C4: Tính toán nhanh hơn

C5:Ta ghi vào thanh công thức "=(E3+F7)*10%"

Chúc bạn thi tốt!!!

mik chắc chắn là powerpoint nhé vì chỉ có microsoft excel, m

Bài 2: Chọn C

Bài 4: 

a: \(\widehat{C}=180^0-80^0-50^0=50^0\)

Xét ΔABC có \(\widehat{A}=\widehat{C}< \widehat{B}\)

nên BC=AB<AC

b: Xét ΔABC có AB<BC<AC

nên \(\widehat{C}< \widehat{A}< \widehat{B}\)

g: \(=\dfrac{x^2+2x-x^2-4x-2x+4}{x\left(x-2\right)\left(x+2\right)}=\dfrac{-4x+4}{x\left(x-2\right)\left(x+2\right)}\)

h: \(=\dfrac{2x^2+1-x^2+1-x^2+x-1}{\left(x+1\right)\left(x^2-x+1\right)}\)

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

\(e,=\dfrac{1}{x-1}-\dfrac{2x}{\left(x^2+1\right)\left(x-1\right)}=\dfrac{x^2-2x+1}{\left(x^2+1\right)\left(x-1\right)}=\dfrac{\left(x-1\right)^2}{\left(x^2+1\right)\left(x-1\right)}=\dfrac{x-1}{x^2+1}\\ f,=\dfrac{3x-1}{2\left(3x+1\right)}+\dfrac{3x+1}{2\left(3x-1\right)}-\dfrac{6x}{\left(3x-1\right)\left(3x+1\right)}\\ =\dfrac{9x^2-6x+1+9x^2+6x+1-12x}{2\left(3x-1\right)\left(3x+1\right)}=\dfrac{2\left(3x-1\right)^2}{2\left(3x-1\right)\left(3x+1\right)}=\dfrac{3x-1}{3x+1}\)

\(g,=\dfrac{x}{x\left(x-2\right)}-\dfrac{x^2+4x}{x\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x\left(x+2\right)}\\ =\dfrac{x^2+2x-x^2-4x-2x+4}{x\left(x-2\right)\left(x+2\right)}=\dfrac{-4x+4}{x\left(x-2\right)\left(x+2\right)}\\ h,=\dfrac{2x^2+1-x^2+1-x^2+x-1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{1}{x^2-x+1}\)

Bài 2 

a, bạn tự vẽ 

b, Hoành độ giao điểm tm pt 

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

\(\Delta'=1-3.2=-5< 0\)

Vậy pt vô nghiệm hay (d) ko cắt (P)

câu hỏi gì kì z

Áp dụng quy tắc Horner , ta có :

Vậy , phaeps chia được thương bằng : x + 1

Cách 2 :

(x² + 2x + 1) : ( x + 1)

(-) ( x + 1)( x + 1) : ( x + 1)

Vậy , thương của phép chia là x + 1