\(\dfrac{\chi+1}{\chi+2}-\dfrac{5}{\chi+2}=\dfrac{12}{\chi^2-4}+1\)
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NT
1
BF
19 tháng 2 2023
`3/4 + 5/6 = 9/12 + 10/12 = 19/12`
`1/2 + 7/12 = 6/12 + 7/12 = 13/12`
`2/3 xx 3/4 = 2/4 = 1/2`
`7/4 : 2 = 7/4 xx 1/2 = 7/8`
VQ
19 tháng 2 2023
\(a,\dfrac{3}{4}+\dfrac{5}{6}=\dfrac{18}{24}+\dfrac{20}{24}=\dfrac{38}{24}=\dfrac{19}{12}\)
\(b,\dfrac{1}{2}+\dfrac{7}{12}=\dfrac{6}{12}+\dfrac{7}{12}=\dfrac{13}{12}\)
\(c,\dfrac{2}{3}x\dfrac{3}{4}=\dfrac{2}{4}\)
\(d,\dfrac{7}{4}:2=\dfrac{7}{4}x\dfrac{1}{2}=\dfrac{7}{8}\)
K
24 tháng 3 2022
\(a,\dfrac{7}{12}+\dfrac{3}{4}\times\dfrac{2}{9}=\dfrac{7}{12}+\dfrac{1}{6}=\dfrac{7}{12}+\dfrac{2}{12}=\dfrac{9}{12}=\dfrac{3}{4}\)
\(b,\dfrac{8}{9}-\dfrac{4}{15}:\dfrac{2}{5}=\dfrac{8}{9}-\dfrac{4}{15}\times\dfrac{5}{2}=\dfrac{8}{9}-\dfrac{2}{3}=\dfrac{8}{9}-\dfrac{6}{9}=\dfrac{2}{9}\)
22 tháng 7 2021
Ta có: \(\left(\dfrac{1}{2}x+\dfrac{3}{4}\right)^2=\dfrac{1}{16}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x+\dfrac{3}{4}=\dfrac{-1}{4}\\\dfrac{1}{2}x+\dfrac{3}{4}=\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=-1\\\dfrac{1}{2}x=\dfrac{-1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)
PT
Phạm Trần Hoàng Anh
CTVHS
22 tháng 7 2021
`(x/2 + 3/4)^2 = 1/16`
`=> (x/2 + 3/4)^2 = (1/4)^2`
Xét `x/2 + 3/4 = 1/4`
`=> x/2 = 1`
`=> x = 2`
Xét `x/2 + 3/4 = -1/4`
`=> x/2 = 1/2`
`=> x = 1`
Vậy `x = 1;2`
(Chúc bạn học tốt)
29 tháng 8 2021
a: Ta có: \(\dfrac{2}{\sqrt{3}+1}+\dfrac{2}{2-\sqrt{3}}\)
\(=\sqrt{3}-1+2+\sqrt{3}\)
\(=2\sqrt{3}+1\)
b: Ta có: \(\dfrac{4}{\sqrt{5}+2}+\dfrac{2}{3+\sqrt{5}}\)
\(=4\sqrt{5}-8+\dfrac{3}{2}-\dfrac{\sqrt{5}}{2}\)
\(=-\dfrac{13}{2}+\dfrac{7}{2}\sqrt{5}\)
8 tháng 3 2022
\(C=\dfrac{-5}{7}+\dfrac{-2}{7}+\dfrac{3}{4}+\dfrac{1}{4}+\dfrac{-1}{5}=-1+1-\dfrac{1}{5}=\dfrac{-1}{5}\)
Sửa đề: \(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)
ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
Ta có: \(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)
\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{12}{\left(x-2\right)\left(x+2\right)}+\dfrac{x^2-4}{\left(x-2\right)\left(x+2\right)}\)
Suy ra: \(x^2+3x+2-5x+10-12-x^2+4=0\)
\(\Leftrightarrow-2x+4=0\)
\(\Leftrightarrow-2x=-4\)
hay x=2(loại)
Vậy: \(S=\varnothing\)
\(\Leftrightarrow\) \(\dfrac{x+1}{x+2}-\dfrac{5}{x+2}-\dfrac{12}{\left(x+2\right)\left(x-2\right)}-1=0\)
\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{12}{\left(x+2\right)\left(x-2\right)}-\dfrac{\left(x+2\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=0\) 0
\(\Leftrightarrow x^2+x-2x-2-5x+10-12-x^2+4=0\)\(\Leftrightarrow\)\(-6x=0\Leftrightarrow x=0\)