\(\dfrac{3}{4}.\left(\dfrac{2}{5}-2x\right)-\dfrac{3}{5}.\left(\dfrac{1}{2}x-\dfrac{5}{4}\right)=\dfrac{9}{20}\)
K
Khách
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Những câu hỏi liên quan
8 tháng 12 2021
\(a,=\dfrac{x^2-20+x^2-7x+10+3x+6}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}=\dfrac{x-2}{x+2}\\ b,=\dfrac{10x+15-4x+6+2x-9}{\left(2x-3\right)\left(2x+3\right)}=\dfrac{4\left(2x+3\right)}{\left(2x-3\right)\left(2x+3\right)}=\dfrac{4}{2x-3}\\ c,=\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}\\ =\dfrac{1}{x}-\dfrac{1}{x+4}=\dfrac{x+4-x}{x\left(x+4\right)}=\dfrac{4}{x\left(x+4\right)}\)
25 tháng 7 2017
a)<=>\(\dfrac{\left(2x-3\right).2}{6}-\dfrac{3.3}{6}=\dfrac{5-2x}{6}-\dfrac{1.3}{6}\)
<=>\(\dfrac{4x-6}{6}-\dfrac{9}{6}=\dfrac{5-2x}{6}-\dfrac{3}{6}\)
<=>\(\dfrac{4x-6}{6}-\dfrac{9}{6}-\dfrac{5-2x}{6}+\dfrac{3}{6}=0\)
<=>\(\dfrac{4x-6-9-5+2x+3}{6}=\dfrac{4x-17}{6}=0\)
<=>\(4x-17=0\)
<=>\(4x=17\)<=>\(x=\dfrac{17}{4}\)
VN
Võ Ngọc Phương
VIP
1 tháng 10 2023
\(\left(3-x\right)^3=-\dfrac{27}{64}\)
\(\left(3-x\right)^3=\left(\dfrac{-3}{4}\right)^3\)
\(=>3-x=\dfrac{-3}{4}\)
\(x=3-\dfrac{-3}{4}=\dfrac{12}{4}+\dfrac{3}{4}\)
\(x=\dfrac{15}{4}\)
________
\(\left(x-5\right)^3=\dfrac{1}{-27}\)
\(\left(x-5\right)^3=\left(\dfrac{-1}{3}\right)^3\)
\(=>x-5=\dfrac{-1}{3}\)
\(x=\dfrac{-1}{3}+5=\dfrac{-1}{3}+\dfrac{15}{3}\)
\(x=\dfrac{14}{3}\)
_____________
\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{27}{8}\)
\(\left(x-\dfrac{1}{2}\right)^3=\left(\dfrac{3}{2}\right)^3\)
\(=>x-\dfrac{1}{2}=\dfrac{3}{2}\)
\(x=\dfrac{3}{2}+\dfrac{1}{2}\)
\(x=2\)
________
\(\left(2x-1\right)^2=\dfrac{1}{4}\)
\(\left(2x-1\right)^2=\left(\dfrac{1}{2}\right)^2\) hoặc \(\left(2x-1\right)^2=\left(\dfrac{-1}{2}\right)^2\)
\(=>2x-1=\dfrac{1}{2}\) \(2x-1=\dfrac{-1}{2}\)
\(2x=\dfrac{1}{2}+1=\dfrac{1}{2}+\dfrac{2}{2}\) \(2x=\dfrac{-1}{2}+1=\dfrac{-1}{2}+\dfrac{2}{2}\)
\(2x=\dfrac{3}{2}\) \(2x=\dfrac{1}{2}\)
\(x=\dfrac{3}{2}:2=\dfrac{3}{2}.\dfrac{1}{2}\) \(x=\dfrac{1}{2}:2=\dfrac{1}{2}.\dfrac{1}{2}\)
\(x=\dfrac{3}{4}\) \(x=\dfrac{1}{4}\)
____________
\(\left(2-3x\right)^2=\dfrac{9}{4}\)
\(\left(2-3x\right)^2=\left(\dfrac{3}{2}\right)^2\) hoặc \(\left(2-3x\right)^2=\left(\dfrac{-3}{2}\right)^2\)
\(=>2-3x=\dfrac{3}{2}\) \(2-3x=\dfrac{-3}{2}\)
\(3x=2-\dfrac{3}{2}=\dfrac{4}{2}-\dfrac{3}{2}\) \(3x=2-\dfrac{-3}{2}=\dfrac{4}{2}+\dfrac{3}{2}\)
\(3x=\dfrac{1}{2}\) \(3x=\dfrac{7}{2}\)
\(x=\dfrac{1}{2}.\dfrac{1}{3}\) \(x=\dfrac{7}{2}.\dfrac{1}{3}\)
\(x=\dfrac{1}{6}\) \(x=\dfrac{7}{6}\)
______________
\(\left(1-\dfrac{2}{3}\right)^2=\dfrac{4}{9}\) -> Kiểm tra đề câu này
1 tháng 10 2023
(3-x)3=(-\(\dfrac{3}{4}\))3
3-x=-\(\dfrac{3}{4}\)
x=3-(-\(\dfrac{3}{4}\))
x=\(\dfrac{15}{4}\)
AT
12 tháng 6 2017
K chép lại đề, lm luôn nhé:
*\(\Rightarrow\) \(\left(\dfrac{7}{2}+2x\right)\cdot\dfrac{8}{3}=\dfrac{16}{3}\)
\(\Rightarrow\dfrac{7}{2}+2x=\dfrac{16}{3}:\dfrac{8}{3}=2\)
\(\Rightarrow2x=2-\dfrac{7}{2}=-\dfrac{3}{2}\)
\(\Rightarrow x=-\dfrac{3}{4}\)
* \(\Rightarrow\left|2x-\dfrac{2}{3}\right|=\dfrac{\dfrac{3}{4}-2}{2}=-\dfrac{5}{8}\)
=> K có gt x nào t/m đề
* Đề sai
* \(\Rightarrow\left[{}\begin{matrix}3x-1=0\\-\dfrac{1}{2}x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=10\end{matrix}\right.\)
*\(\Rightarrow\dfrac{1}{3}:\left(2x-1\right)=-5-\dfrac{1}{4}=-\dfrac{21}{4}\)
\(\Rightarrow2x-1=\dfrac{1}{3}:\left(-\dfrac{21}{4}\right)=-\dfrac{4}{63}\)
\(\Rightarrow2x=-\dfrac{4}{63}+1=\dfrac{59}{63}\)
\(\Rightarrow x=\dfrac{59}{63}:2=\dfrac{59}{126}\)
* \(\Rightarrow\left(2x+\dfrac{3}{5}\right)^2=\dfrac{9}{25}\)
\(\Rightarrow\left[{}\begin{matrix}2x+\dfrac{3}{5}=\dfrac{3}{5}\\2x+\dfrac{3}{5}=-\dfrac{3}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=0\Rightarrow x=0\\2x=-\dfrac{6}{5}\Rightarrow x=-\dfrac{3}{5}\end{matrix}\right.\)
* \(\Rightarrow-5x-1-\dfrac{1}{2}x+\dfrac{1}{3}=\dfrac{3}{2}x-\dfrac{5}{6}\)
\(\Rightarrow-5x-\dfrac{1}{2}x-\dfrac{3}{2}x=-\dfrac{5}{6}+1-\dfrac{1}{3}\)
\(\Rightarrow-7x=-\dfrac{1}{6}\)
\(\Rightarrow x=-\dfrac{1}{6}:7=-\dfrac{1}{42}\)
12 tháng 6 2017
a)\(\left(3\dfrac{1}{2}+2x\right).2\dfrac{2}{3}=5\dfrac{1}{3}\)
\(\left(\dfrac{7}{2}+2x\right).\dfrac{8}{3}=\dfrac{16}{3}\)
\(\dfrac{7}{2}+2x=\dfrac{16}{3}:\dfrac{8}{3}=2\)
\(2x=2-\dfrac{7}{2}=\dfrac{-3}{2}\Rightarrow x=\dfrac{-3}{4}\)
b)\(\dfrac{3}{4}-2.\left|2x-\dfrac{2}{3}\right|=2\)
\(2.\left|2x-\dfrac{2}{3}\right|=\dfrac{3}{4}-2=\dfrac{-1}{4}\)
\(\Rightarrow\left|2x-3\right|=\dfrac{-1}{8}\)
\(\Rightarrow x\in\varnothing\)
c) Đề sai,bạn có viết chữ x đâu,đó là phép tính mà.
d)\(\left(3x-1\right)\left(\dfrac{-1}{2}x+5\right)=0\)
\(\Leftrightarrow3x-1=0\Rightarrow x=\dfrac{1}{3}\)
\(\Leftrightarrow\dfrac{-1}{2}x+5=0\Rightarrow x=10\)
e)\(\dfrac{1}{4}+\dfrac{1}{3}:\left(2x-1\right)=-5\)
\(\dfrac{1}{3}:\left(2x-1\right)=-5-\dfrac{1}{4}=\dfrac{-21}{4}\)
\(2x-1=\dfrac{1}{3}:\dfrac{-21}{4}=\dfrac{-4}{63}\)
\(\Rightarrow2x=\dfrac{59}{63}\Rightarrow x=\dfrac{59}{126}\)
g)\(\left(2x+\dfrac{3}{5}\right)^2-\dfrac{9}{25}=0\)
\(\left(2x+\dfrac{3}{5}\right)^2=0+\dfrac{9}{25}=\dfrac{9}{25}\)
\(\dfrac{9}{25}=\left(\dfrac{3}{5}\right)^2=\left(\dfrac{-3}{5}\right)^2\)
\(th1:x=0\)
\(th2:x=\dfrac{-3}{5}\)
h)\(-5\left(x+\dfrac{1}{5}\right)-\dfrac{1}{2}\left(x-\dfrac{2}{3}\right)=\dfrac{3}{2}x-\dfrac{5}{6}\)
\(-5x+-1-\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{3}{2}x-\dfrac{5}{6}\)
\(\Leftrightarrow-5x+-1+\dfrac{5}{6}-\dfrac{1}{3}=2x\)
\(-5x+\dfrac{-1}{2}=2x\)
\(\dfrac{-1}{2}=2x+5x\)
\(\dfrac{-1}{2}=7x\Rightarrow x=\dfrac{-1}{14}\)
16 tháng 2 2021
a) Ta có: \(\left(2x-3\right)^2=\left(2x-3\right)\left(x+1\right)\)
\(\Leftrightarrow\left(2x-3\right)^2-\left(2x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(2x-3-x-1\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=4\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{3}{2};4\right\}\)
b) Ta có: \(x\left(2x-9\right)=3x\left(x-5\right)\)
\(\Leftrightarrow x\left(2x-9\right)-3x\left(x-5\right)=0\)
\(\Leftrightarrow x\left(2x-9\right)-x\left(3x-15\right)=0\)
\(\Leftrightarrow x\left(2x-9-3x+15\right)=0\)
\(\Leftrightarrow x\left(6-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\6-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
Vậy: S={0;6}
c) Ta có: \(3x-15=2x\left(x-5\right)\)
\(\Leftrightarrow3\left(x-5\right)-2x\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(3-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\3-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\2x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{5;\dfrac{3}{2}\right\}\)
d) Ta có: \(\dfrac{5-x}{2}=\dfrac{3x-4}{6}\)
\(\Leftrightarrow6\left(5-x\right)=2\left(3x-4\right)\)
\(\Leftrightarrow30-6x=6x-8\)
\(\Leftrightarrow30-6x-6x+8=0\)
\(\Leftrightarrow-12x+38=0\)
\(\Leftrightarrow-12x=-38\)
\(\Leftrightarrow x=\dfrac{19}{6}\)
Vậy: \(S=\left\{\dfrac{19}{6}\right\}\)
e) Ta có: \(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)
\(\Leftrightarrow\dfrac{3\left(3x+2\right)}{6}-\dfrac{3x+1}{6}=\dfrac{12x}{6}+\dfrac{10}{6}\)
\(\Leftrightarrow6x+4-3x-1=12x+10\)
\(\Leftrightarrow3x+3-12x-10=0\)
\(\Leftrightarrow-9x-7=0\)
\(\Leftrightarrow-9x=7\)
\(\Leftrightarrow x=-\dfrac{7}{9}\)
Vậy: \(S=\left\{-\dfrac{7}{9}\right\}\)
10 tháng 5 2022
1: \(\left(3x-\dfrac{1}{5}\right)^2=\left(-\dfrac{3}{25}\right)^2\)
=>3x-1/5=3/25 hoặc 3x-1/5=-3/25
=>3x=8/25 hoặc 3x=2/25
=>x=8/75 hoặc x=2/75
2: \(\left(2x-\dfrac{1}{3}\right)^2=\left(-\dfrac{2}{9}\right)^2\)
=>2x-1/3=2/9 hoặc 2x-1/3=-2/9
=>2x=5/9 hoặc 2x=1/9
=>x=5/18 hoặc x=1/18
29 tháng 10 2021
3: \(\left|x-\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)
\(\Leftrightarrow\left|x-\dfrac{3}{4}\right|=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{1}{2}\\x-\dfrac{3}{4}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=\dfrac{1}{4}\end{matrix}\right.\)
YN
4 tháng 7 2023
Để giải phương trình, ta sẽ thực hiện các bước sau: Bước 1: Giải các phép tính trong phương trình. 32x^(-1) + 2.9x^(-1) = 405(13)^(-1) + 5.(13)^2 + 1 = 1493(31)^(-1) + 5.(31)^2 + 1 = 9314(35)^(-1) Bước 2: Rút gọn các số hạng. 32x^(-1) + 2.9x^(-1) = 405/13 + 5.(13)^2 + 1 = 1493/31 + 5.(31)^2 + 1 = 9314/35 Bước 3: Đưa các số hạng về cùng mẫu số. 32x^(-1) + 2.9x^(-1) = (405/13).(31/31) + 5.(13)^2 + 1 = (1493/31).(13/13) + 5.(31)^2 + 1 = 9314/35 Bước 4: Tính toán các số hạng. 32x^(-1) + 2.9x^(-1) = 405.(31)/13.(31) + 5.(13)^2 + 1 = 1493.(13)/31.(13) + 5.(31)^2 + 1 = 9314/35 Bước 5: Tính tổng các số hạng. 32x^(-1) + 2.9x^(-1) = 405.(31)/403 + 5.(13)^2 + 1 = 1493.(13)/403 + 5.(31)^2 + 1 = 9314/35 Bước 6: Đưa phương trình về dạng chuẩn. 32x^(-1) + 2.9x^(-1) - 9314/35 = 0 Bước 7: Giải phương trình. Để giải phương trình này, ta cần biến đổi nó về dạng tương đương. Nhân cả hai vế của phương trình với 35 để loại bỏ mẫu số. 35.(32x^(-1) + 2.9x^(-1) - 9314/35) = 0 1120x^(-1) + 101.5x^(-1) - 9314 = 0 Bước 8: Tìm giá trị của x. Để tìm giá trị của x, ta cần giải phương trình này. Tuy nhiên, phương trình này không thể giải được vì x có mũ là -1.
Đề yêu cầu gì vậy em?