Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
f: =-1/8-7/6+3/4-1
=-3/24-28/24+18/24-1
=-31/24+18/24-1
=-13/24-1=-37/24
g: \(=6\cdot\dfrac{-8}{27}-3\cdot\dfrac{4}{9}+\dfrac{4}{3}+4\)
=-48/27+4
=108/27-48/27
=60/27
=20/9
h: \(=\left[6\cdot\dfrac{1}{9}+1+1\right]\cdot\left(-3\right)-1\)
=(2/3+2)*(-3)-1
=-2-6-1
=-3-6=-9
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 \(\widehat{C}< \widehat{A}< \widehat{B}\)
Bài 2 :
a,\(\frac{x-1}{3}=2-\frac{x}{-2}\)
\(\Leftrightarrow\frac{x-1}{3}=\frac{-4-x}{-2}\Leftrightarrow-2x+2=-12-3x\Leftrightarrow x=-14\)
b, \(\frac{x-1}{x+5}=\frac{6}{7}\Leftrightarrow7x-7=6x+30\Leftrightarrow x=37\)
c, \(\frac{2x-1}{4}=\frac{4}{2x-1}\Leftrightarrow\left(2x-1\right)^2=16\)
\(\Leftrightarrow\left(2x-1\right)^2-4^2=0\Leftrightarrow\left(2x-5\right)\left(2x+3\right)=0\Leftrightarrow x=\frac{5}{2};-\frac{3}{2}\)
\(a,x+\dfrac{1}{2}=\dfrac{3}{4}\\ x=\dfrac{3}{4}-\dfrac{1}{2}\\ x=\dfrac{1}{2}\\ b,-\dfrac{2}{3}-x=1\\x=-\dfrac{2}{3}-1\\ x=-\dfrac{5}{3}\\ d,\dfrac{1}{4}+\dfrac{3}{4}:x=\dfrac{5}{2}\\ \dfrac{3}{4}:x=\dfrac{5}{2}-\dfrac{1}{4}\\ \dfrac{3}{4}:x=\dfrac{9}{4}\\ x=\dfrac{3}{4}:\dfrac{9}{4}\\ x=\dfrac{1}{3}\\ e,\left(x+\dfrac{1}{4}\right)\cdot\dfrac{3}{4}=-\dfrac{5}{8}\\ x+\dfrac{1}{4}=-\dfrac{5}{8}:\dfrac{3}{4}\\ x+\dfrac{1}{4}=\dfrac{5}{6}\\ x=\dfrac{5}{6}-\dfrac{1}{4}\\ x=\dfrac{7}{12}\)
\(g,\dfrac{x-3}{15}=\dfrac{-2}{5}\\ 5\left(x-3\right)=-30\\ x-3=-6\\ x=-6+3\\ x=-3\\ h,\dfrac{x}{-2}=\dfrac{-8}{x}\\ x^2=16\\ x=\pm\sqrt{16}\\ x=\pm4\\ k,\dfrac{x+2}{3}=\dfrac{x-4}{5}\\ 5\left(x+2\right)=3\left(x-4\right)\\ 5x+10=3x-12\\ 5x-3x=-12-10\\ 2x=-22\\ x=-11\)
\(m,\left(2x-1\right)^2=4\\ \Rightarrow\left[{}\begin{matrix}2x-1=2\\2x-1=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=3\\2x=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
đề bài là j
nghiệm của đa thức 1 biến ,ko thấy à