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1
Những câu hỏi liên quan
NT
1
NV
Nguyễn Việt Lâm
Giáo viên
27 tháng 7 2021
2.1
ĐKXĐ: \(x\ge-\dfrac{1}{16}\)
\(x^2-x-20-2\left(\sqrt{16x+1}-9\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+4\right)-\dfrac{32\left(x-5\right)}{\sqrt{16x+1}+9}=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+4-\dfrac{32}{\sqrt{16x+1}+9}\right)=0\) (1)
Do \(x\ge-\dfrac{1}{16}\Rightarrow\left\{{}\begin{matrix}\dfrac{32}{\sqrt{16x+1}+9}< \dfrac{32}{9}\\x+4\ge-\dfrac{1}{16}+4=\dfrac{63}{16}>\dfrac{32}{9}\end{matrix}\right.\)
\(\Rightarrow x+4-\dfrac{32}{\sqrt{16x+1}+9}>0\)
Nên (1) tương đương:
\(x-5=0\)
\(\Leftrightarrow x=5\)
Câu 2.2, 2.3 đề lỗi không dịch được
NT
3
NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 7 2021
2.2 Đề lỗi không dịch được
2.3
\(\Delta'=4m^2-2\left(2m^2-1\right)=2>0\Rightarrow\) pt luôn có 2 nghiệm pb
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=2m\\x_1x_2=\dfrac{2m^2-1}{2}\end{matrix}\right.\)
\(\Rightarrow\left(x_1-x_2\right)^2=\left(x_1+x_2\right)^2-4x_1x_2=2\Rightarrow\left[{}\begin{matrix}x_1-x_2=\sqrt{2}\\x_1-x_2=-\sqrt{2}\end{matrix}\right.\)
Do \(x_1\) là nghiệm của pt nên ta có:
\(2x_1^2-4mx_1+2m^2-1=0\Rightarrow2x_1^2=4mx_1-2m^2+1\)
Thế vào bài toán:
\(4mx_1-2m^2+1-4mx_2+2m^2-9< 0\)
\(\Leftrightarrow m\left(x_1-x_2\right)< 2\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2}m< 2\\-\sqrt{2}m< 2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}m< \sqrt{2}\\m>-\sqrt{2}\end{matrix}\right.\)
(Bản thân câu này đề bài cũng rất dở)
9 tháng 12 2021
\(P=\dfrac{x^3+8y^3}{4^3+4^3}=\dfrac{\left(x+2y\right)^3-3\cdot x\cdot2y\cdot\left(x+2y\right)}{128}\)
\(=\dfrac{\left(-8\right)^3-6\cdot\left(-6\right)\cdot\left(-8\right)}{128}=\dfrac{128-6\cdot48}{128}=-\dfrac{5}{4}\)
23 tháng 12 2021
a: \(M=\dfrac{5x+5-8-4x+4}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}=\dfrac{1}{x-1}\)
b: \(N=\dfrac{5x+5-8-4x+4}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}=\dfrac{1}{x-1}\)
23 tháng 12 2021
1. \(M=\dfrac{5}{x-1}-\dfrac{8}{x^2-1}-\dfrac{4}{x+1}\left(x\ne\pm1\right).\)
\(M=\dfrac{5\left(x+1\right)-8-4\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\)\(M=\dfrac{5x+5-8-4x+4}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}\)
\(M=\dfrac{1}{x-1}.\)
2. \(N=\dfrac{5}{x-1}+\dfrac{8}{1-x^2}-\dfrac{4}{x+1}\left(x\ne\pm1\right).\)
\(N=\dfrac{5\left(x+1\right)-8-4\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{5x+5-8-4x+4}{\left(x-1\right)\left(x+1\right)}\)
\(N=\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}=\dfrac{1}{x-1}.\)
3. \(Q=\dfrac{1}{2x-1}-\dfrac{4}{4x^2-1}-\dfrac{2}{2x+1}\left(x\ne\pm\dfrac{1}{2}\right).\)
\(Q=\dfrac{2x+1-4-2\left(2x-1\right)}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{2x-3-4x+2}{\left(2x-1\right)\left(2x+1\right)}\)
\(Q=\dfrac{-2x-1}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{-1}{2x-1}.\)
4. \(F=\dfrac{x+3}{x-2}+\dfrac{x+2}{3-x}+\dfrac{x+2}{x^2-5x+6}\left(x\ne2,x\ne3\right).\)
\(F=\dfrac{x+3}{x-2}-\dfrac{x+2}{x-3}+\dfrac{x+2}{\left(x-3\right)\left(x-2\right)}\)
\(F=\dfrac{\left(x+3\right)\left(x-3\right)-\left(x+2\right)\left(x-2\right)+x+2}{\left(x-2\right)\left(x-3\right)}\)
\(F=\dfrac{x^2-9-x^2+4+x+2}{\left(x-2\right)\left(x-3\right)}=\dfrac{x-3}{\left(x-2\right)\left(x-3\right)}\)
\(F=\dfrac{1}{x-2}.\)
giúp em với ạ e, đang cần gấp lắm ạ
H
2 tháng 7 2023
\(b,\dfrac{1}{2}+\dfrac{13}{19}-\dfrac{4}{9}+\dfrac{6}{19}+\dfrac{5}{18}\\ =\left(\dfrac{1}{2}+\dfrac{5}{18}\right)+\left(\dfrac{13}{19}+\dfrac{6}{19}\right)-\dfrac{4}{9}\\ =\left(\dfrac{9}{18}+\dfrac{5}{18}\right)+\dfrac{19}{19}-\dfrac{4}{9}\\ =\dfrac{14}{18}+1-\dfrac{4}{9}\\ =\dfrac{7}{9}+1-\dfrac{4}{9}\\ =\left(\dfrac{7}{9}-\dfrac{4}{9}\right)+1\\ =\dfrac{3}{9}+1\\ =\dfrac{1}{3}+1\\ =\dfrac{4}{3}\)
\(c,\dfrac{-20}{23}+\dfrac{2}{3}-\dfrac{3}{23}+\dfrac{2}{5}+\dfrac{7}{15}\\ =\left(-\dfrac{20}{23}-\dfrac{3}{23}\right)+\left(\dfrac{2}{5}+\dfrac{7}{15}\right)+\dfrac{2}{3}\\ =-\dfrac{23}{23}+\left(\dfrac{6}{15}+\dfrac{7}{15}\right)+\dfrac{2}{3}\\ =-1+\dfrac{13}{15}+\dfrac{2}{3}\\ =-\dfrac{15}{15}+\dfrac{13}{15}+\dfrac{10}{15}\\ =\dfrac{8}{15}\)
\(e,\dfrac{5}{7}.\dfrac{5}{11}+\dfrac{5}{7}.\dfrac{2}{11}-\dfrac{5}{7}.\dfrac{14}{11}\\ =\dfrac{5}{7}.\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)\\ =\dfrac{5}{7}.\dfrac{-7}{11}\\ =-\dfrac{35}{77}\\ =-\dfrac{5}{11}\)
\(f,\dfrac{2}{11}.\dfrac{-5}{4}+\dfrac{-9}{11}.\dfrac{5}{4}+1\dfrac{3}{4}\\ =-\dfrac{2}{11}.\dfrac{5}{4}+\dfrac{-9}{11}.\dfrac{5}{4}+\dfrac{7}{4}\\=\dfrac{5}{4}.\left(-\dfrac{2}{11}+\dfrac{-9}{11}\right)+\dfrac{7}{4}\\ =\dfrac{5}{4}.1+\dfrac{7}{4}\\ =\dfrac{5}{4}+\dfrac{7}{4}\\=\dfrac{12}{4}\\ =3\)
\(h,\dfrac{7}{4}\cdot\dfrac{29}{5}-\dfrac{7}{5}\cdot\dfrac{9}{4}+3\dfrac{2}{13}\\ =\dfrac{7}{4}\cdot\dfrac{29}{5}-\dfrac{7}{4}\cdot\dfrac{9}{5}+\dfrac{41}{13}\\ =\dfrac{7}{4}\cdot\left(\dfrac{29}{5}-\dfrac{9}{5}\right)+\dfrac{41}{13}\\ =\dfrac{7}{4}\cdot\dfrac{20}{5}+\dfrac{41}{13}\\ =\dfrac{7}{4}.4+\dfrac{41}{13}\\ =\dfrac{28}{4}+\dfrac{41}{13}\\ =7+\dfrac{41}{13}\\ =\dfrac{132}{13}\)
T
2
19 tháng 7 2021
b) Ta có: \(\sqrt{150}-\sqrt{1.6}\cdot\sqrt{60}+4.5\cdot\sqrt{2\dfrac{2}{3}}-\sqrt{6}\)
\(=5\sqrt{6}-4\sqrt{6}-\sqrt{6}+\dfrac{9}{2}\cdot\sqrt{\dfrac{8}{3}}\)
\(=\dfrac{9}{2}\cdot\dfrac{2\sqrt{2}}{\sqrt{3}}\)
\(=3\sqrt{6}\)
LN
19 tháng 7 2021
\(\sqrt{150}+\sqrt{1,6}.\sqrt{60}+4.5\sqrt{2\dfrac{2}{3}}-\sqrt{6}\\ =5\sqrt{6}+4\sqrt{6}+3\sqrt{6}-\sqrt{6}\\ =11\sqrt{6}\)
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