tìm x biết x+\(\dfrac{2}{7}\)=\(\dfrac{-49}{x+2^2}\)
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P
1
20 tháng 1 2022
\(\Leftrightarrow\left(x+2\right)^3=-343\)
=>x+2=-7
hay x=-9
26 tháng 9 2021
a: Ta có: \(\dfrac{x+1}{2}=\dfrac{2}{x+1}\)
\(\Leftrightarrow\left(x+1\right)^2=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=2\\x+1=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)
b: Ta có: \(\dfrac{\left(x-2\right)^2}{7}=\dfrac{49}{\left(x-2\right)}\)
\(\Leftrightarrow x-2=7\)
hay x=9
7 tháng 10 2021
\(x:\left[\dfrac{8}{5}\cdot\left(\dfrac{2}{3}\right)^2-\dfrac{2}{5}\right]=\dfrac{15}{7}+\dfrac{6}{5}\left[\left(2\dfrac{1}{7}\right)^2-\dfrac{50}{49}\right]\)
\(\Leftrightarrow x:\left[\dfrac{32}{45}-\dfrac{18}{45}\right]=\dfrac{15}{7}+\dfrac{6}{5}\cdot\left(\dfrac{225}{49}-\dfrac{50}{49}\right)\)
\(\Leftrightarrow x:\dfrac{14}{45}=\dfrac{15}{7}+\dfrac{6}{5}\cdot\dfrac{25}{7}\)
\(\Leftrightarrow x:\dfrac{14}{45}=\dfrac{45}{7}\)
\(\Leftrightarrow x=2\)
6 tháng 7 2023
a: Sửa đề; \(A=\dfrac{7.2:2\cdot28.6+1.43\cdot2\cdot64}{1+3+5+7+...+49-339}\)
\(=\dfrac{3.6\cdot28.6+2.86\cdot64}{1+3+5+...+49-339}\)
\(=\dfrac{2.86\left(64+36\right)}{25^2-339}=\dfrac{286}{286}=1\)
b: =>2(x+7/8)=6*13/4=78/4=39/2
=>x+7/8=39/4
=>x=71/8
T
9 tháng 8 2023
a) \(\dfrac{49}{81}=\dfrac{7^x}{9^x}\)(sửa đề)
\(\Leftrightarrow\left(\dfrac{7}{9}\right)^2=\left(\dfrac{7}{9}\right)^x\)\(\Rightarrow x=2\)
b) \(\dfrac{-64}{343}=\left(-\dfrac{4^x}{7^x}\right)\)(sửa đề)
\(\Leftrightarrow\left(-\dfrac{4}{7}\right)^3=\left(-\dfrac{4}{7}\right)^x\) \(\Rightarrow x=3\)
c) \(\dfrac{9}{144}=\dfrac{3^x}{12^x}\)(sửa đề)
\(\Leftrightarrow\left(\dfrac{3}{12}\right)^2=\left(\dfrac{3}{12}\right)^x\Rightarrow x=2\)
d) \(-\dfrac{1}{32}=\left(-\dfrac{1^x}{2^x}\right)\)(sửa đề)
\(\Leftrightarrow\left(-\dfrac{1}{2}\right)^5=\left(-\dfrac{1}{2}\right)^x\Rightarrow x=5\)
Mong bạn xem lại đề bài.
MC
27 tháng 12 2020
a, \(\dfrac{x^2-49}{x-7}\) + x - 2 = \(\dfrac{\left(x-7\right)\left(x+7\right)}{x-7}\) + x - 2 = x + 7 + x - 2 = 2x + 5
b, \(\left(\dfrac{x}{x^2-36}-\dfrac{x-6}{x^2+6x}\right)\) . \(\dfrac{x^2+6x}{2x-6}\)
= \(\left(\dfrac{x^2}{x\left(x-6\right)\left(x+6\right)}-\dfrac{\left(x-6\right)^2}{x\left(x+6\right)\left(x-6\right)}\right)\) . \(\dfrac{x\left(x+6\right)}{2x-6}\)
= \(\left(\dfrac{x^2-\left(x-6\right)^2}{x\left(x-6\right)\left(x+6\right)}\right)\) . \(\dfrac{x\left(x+6\right)}{2x-6}\)
= \(\left(\dfrac{6\left(2x-6\right)}{x\left(x-6\right)\left(x+6\right)}\right)\) . \(\dfrac{x\left(x+6\right)}{2x-6}\)
= \(\dfrac{6}{x-6}\)
BT
27 tháng 12 2020
1. = \(\dfrac{\left(x-7\right)\left(x+7\right)}{x-7}\) + x-2
= x+7 +x-2
= 2x-5
2. = (\(\dfrac{x}{\left(x-6\right)\left(x+6\right)}\) - \(\dfrac{x-6}{x\left(x+6\right)}\) ) \(^{\dfrac{x^2+6x}{2x-6}}\)
= ( \(\dfrac{x^2}{x\left(x-6\right)\left(x+6\right)}\) - \(\dfrac{\left(x-6\right)\left(x-6\right)}{x\left(x-6\right)\left(x+6\right)}\) ) \(\dfrac{x^2+6x}{2x-6}\)
= \(\dfrac{x^2-\left(x^2-12x+36\right)}{x\left(x-6\right)\left(x+6\right)}\) . \(\dfrac{x^2+6x}{2x-6}\)
= \(\dfrac{x^2-x^2+12x-36}{x\left(x-6\right)\left(x+6\right)}\) . \(\dfrac{x^2+6x}{2x-6}\)
= \(\dfrac{12x-36}{x\left(x-6\right)\left(x+6\right)}\) . \(\dfrac{x^2+6x}{2x-6}\)
= \(\dfrac{12\left(x-3\right)x\left(x+6\right)}{x\left(x-6\right)\left(x+6\right)2\left(x-3\right)}\)
= \(\dfrac{6}{x-6}\)
Chúc bạn học tốt!
TM
24 tháng 4 2022
a) \(\dfrac{3}{x-7}+\dfrac{2}{x+7}=\dfrac{5}{x^2-49}\)
(ĐKXĐ: x khác 7; x khác -7)
<=>\(\dfrac{3.\left(x+7\right)}{\left(x-7\right).\left(x+7\right)}+\dfrac{2.\left(x-7\right)}{\left(x+7\right).\left(x-7\right)}=\dfrac{5}{\left(x+7\right).\left(x-7\right)}\)
=> 3x + 21 + 2x - 14 = 5
<=> 3x + 2x = 5 + 14 - 21
<=> 5x = -2
<=> x = \(\dfrac{-2}{5}\)
Vậy S = { \(\dfrac{-2}{5}\) }
TM
24 tháng 4 2022
b) \(\dfrac{2x-1}{3}-\dfrac{x+3}{2}>1+\dfrac{5x}{6}\)
<=> \(\dfrac{2.\left(2x-1\right)}{3.2}-\dfrac{3.\left(x+3\right)}{3.2}>\dfrac{1.6}{6}+\dfrac{5x}{6}\)
=> 4x - 2 - 3x - 9 > 6 + 5x
<=> 4x - 3x - 5x > 6 + 9 + 2
<=> -4x > 17
<=> \(\dfrac{-17}{4}\)
Vậy S = { \(\dfrac{-17}{4}\) }
HL
1
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 12 2022
Do \(\left|x-\dfrac{2}{3}\right|\ge0;\forall x\)
Mà \(-\dfrac{26}{\sqrt{81}}< 0\)
\(\Rightarrow\) Không tồn tại x để \(\left|x-\dfrac{2}{3}\right|< -\dfrac{26}{\sqrt{81}}\)
Hay ko tồn tại số nguyên x thỏa mãn đề bài
9 tháng 12 2022
2: \(\left(\dfrac{7}{a+7}+\dfrac{a^2+49}{a^2-49}-\dfrac{7}{a-7}\right):\dfrac{a+1}{2}\)
\(=\dfrac{7a-49+a^2+49-7a-49}{\left(a-7\right)\left(a+7\right)}\cdot\dfrac{2}{a+1}\)
\(=\dfrac{a^2-49}{\left(a-7\right)\left(a+7\right)}\cdot\dfrac{2}{a+1}=\dfrac{2}{a+1}\)
3: \(=\dfrac{x^4-4x^2+4x^2}{x^2-4}\cdot\left(\dfrac{x+2}{x-4}+\dfrac{2-3x}{x\left(x^2-4\right)}\cdot\dfrac{x^2-4}{x-2}\right)\)
\(=\dfrac{x^4}{\left(x-2\right)\left(x+2\right)}\cdot\left(\dfrac{x+2}{x-4}+\dfrac{2-3x}{x\left(x-2\right)}\right)\)
\(=\dfrac{x^4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x\left(x^2-4\right)+\left(2-3x\right)\left(x-4\right)}{x\left(x-2\right)\left(x-4\right)}\)
\(=\dfrac{x^4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x^3-4x+2x-8-3x^2+12x}{x\left(x-2\right)\left(x-4\right)}\)
\(=\dfrac{x^4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x^3-3x^2+10x-8}{x\left(x-2\right)\left(x-4\right)}\)
\(=\dfrac{x^4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x^3-x^2-2x^2+2x+8x-8}{x\left(x-2\right)\left(x-4\right)}\)
\(=\dfrac{x^3\left(x-1\right)\left(x^2-2x+8\right)}{\left(x-2\right)^2\cdot\left(x+2\right)\left(x-4\right)}\)
\(x+\dfrac{2}{7}=\dfrac{-49}{x+2^2}\)
\(⇔x+\dfrac{2}{7}+\dfrac{49}{x+4}=0\)
\(⇔\dfrac{7x(x+4)+14(x+4)+343}{7(x+4)}=0\)
\(⇔\dfrac{7x^2+28x+14x+48+343}{7(x+4)}=0\)
\(⇔\dfrac{7x^2+42x+391}{7(x+4)}=0\)
\(⇔7x^2+42x+391=0\)
\(⇔x∉R\)