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19 tháng 9 2019

\(\frac{1-5x}{2}=\frac{13}{36}-\frac{x+1}{3}\)

\(\Leftrightarrow\frac{1-5x}{2}=\frac{13}{36}-\frac{12.\left(x+1\right)}{36}\)

\(\Leftrightarrow\frac{1-5x}{2}=\frac{13-12x-12}{36}\)

\(\Leftrightarrow\frac{1-5x}{2}=\frac{1-12x}{36}\)

\(\Leftrightarrow2.\left(1-12x\right)=36.\left(1-5x\right)\)

\(\Leftrightarrow2-24x=36-180x\)

\(\Leftrightarrow-24x+180x=36-2\)

\(\Leftrightarrow156x=34\)

\(\Leftrightarrow x=\frac{17}{78}\)

13 tháng 11 2019

@Nguyễn Việt Lâm em sắp ktra, anh giúp em bài này với ạ ....

13 tháng 11 2019

Akai Haruma giúp em giải phương trình trên được ko ạ ^_^

26 tháng 9 2016

\(\left(\frac{\frac{17}{24}.9\frac{1}{2}-3\frac{1}{4}.\frac{17}{24}}{3\frac{1}{2}.2\frac{13}{36}+2\frac{13}{36}.2\frac{3}{4}}-\frac{1}{2}\right)^{-2}\)

\(=\left(\frac{\frac{17}{24}.\left(9\frac{1}{2}-3\frac{1}{4}\right)}{2\frac{13}{36}.\left(3\frac{1}{2}+2\frac{3}{4}\right)}-\frac{1}{2}\right)^{-2}\)

\(=\left(\frac{\frac{17}{24}.\left(\frac{19}{2}-\frac{13}{4}\right)}{\frac{85}{36}.\left(\frac{7}{2}+\frac{11}{4}\right)}-\frac{1}{2}\right)^{-2}\)

\(=\left(\frac{\frac{17}{24}.\frac{19.2-13}{4}}{\frac{85}{36}.\frac{7.2+11}{4}}-\frac{1}{2}\right)^{-2}\)

\(=\left(\frac{\frac{17}{24}.\frac{25}{4}}{\frac{85}{36}.\frac{25}{4}}-\frac{1}{2}\right)^{-2}\)

\(=\left(\frac{17}{24}:\frac{85}{36}-\frac{1}{2}\right)^{-2}\)

\(=\left(\frac{17}{24}.\frac{36}{85}-\frac{1}{2}\right)^{-2}\)

\(=\left(\frac{3}{10}-\frac{1}{2}\right)^{-2}\)

\(=\left(\frac{3-5}{10}\right)^{-2}\)

\(=\left(\frac{-1}{5}\right)^{-2}\)

\(=\frac{1}{\left(-\frac{1}{5}\right)^2}=\frac{1}{\frac{\left(-1\right)^2}{5^2}}=\frac{1}{\frac{1}{25}}=25\)

AH
Akai Haruma
Giáo viên
1 tháng 12 2019

Lời giải:

Áp dụng BĐT AM-GM ta có:

\(4x^2+1\geq 4x\)

\(\Rightarrow \left\{\begin{matrix} 5x^2-x+3\geq x^2+3x+2\\ 5x^2+x+\geq x^2+5x+6\\ 5x^2+3x+13\geq x^2+7x+12\\ 5x^2+5x+21\geq x^2+9x+20\end{matrix}\right.\)

\(\text{VT}\leq \frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+\frac{1}{x^2+7x+12}+\frac{1}{x^2+9x+20}\)

\(\Leftrightarrow \text{VT}\leq \frac{1}{(x+1)(x+2)}+\frac{1}{(x+2)(x+3)}+\frac{1}{(x+3)(x+4)}+\frac{1}{(x+4)(x+5)}\)

\(\Leftrightarrow \text{VT}\leq \frac{(x+2)-(x+1)}{(x+1)(x+2)}+\frac{(x+3)-(x+2)}{(x+2)(x+3)}+\frac{(x+4)-(x+3)}{(x+3)(x+4)}+\frac{(x+5)-(x+4)}{(x+4)(x+5)}\)

\(\Leftrightarrow \text{VT}\leq \frac{1}{x+1}-\frac{1}{x+5}\)

\(\Leftrightarrow \text{VT}\leq \frac{4}{x^2+6x+5}\)

Dấu "=" xảy ra khi $4x^2=1, x>0$ hay $x=\frac{1}{2}$

Vậy $x=\frac{1}{2}$ là nghiệm của PT.

30 tháng 11 2019

Nguyễn Việt Lâm anh giúp em pt trên với ạ !!!

4 tháng 2 2020

a/ \(7x-5=13-5x\)

\(\Leftrightarrow7x+5x=13+5\)

\(\Leftrightarrow12x=18\)

\(\Leftrightarrow x=\frac{3}{2}\)

b/\(5\left(2x-3\right)-4\left(5x-7\right)=19-2\left(x+11\right)\)

\(\Leftrightarrow10x-15-20x+28=19-2x-22\)

\(\Leftrightarrow10x-20x+2x=19-22-28+15\)

\(\Leftrightarrow-8x=-16\)

\(\Leftrightarrow x=2\)

c/ \(\frac{2x-1}{3}-\frac{5x+2}{7}=x+13\)

\(\Leftrightarrow\frac{7\left(2x-1\right)-3\left(5x+2\right)-21\left(x+13\right)}{21}=0\)

\(\Leftrightarrow14x-7-15x-6-21x-273=0\)

\(\Leftrightarrow-22x-286=0\)

\(\Leftrightarrow x=-13\)

e/ \(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{\left(x+1\right)\left(x+2\right)}\)

\(\Leftrightarrow\frac{2}{x+1}-\frac{1}{x-2}-\frac{3x-11}{\left(x+1\right)\left(x+2\right)}=0\)

\(\Leftrightarrow\frac{2\left(x-2\right)\left(x+2\right)-\left(x+1\right)\left(x+2\right)-\left(3x-11\right)\left(x-2\right)}{\left(x+1\right)\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow\frac{2\left(x^2-4\right)-\left(x^2+3x+2\right)-\left(3x^2-17x+22\right)}{\left(x+1\right)\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow2x^2-8-x^2-3x-2-3x^2+17x-22=0\)

\(\Leftrightarrow-2x^2+14x-32=0\)

\(\Leftrightarrow x^2-7x+16=0\)

\(\Leftrightarrow x=\frac{-\left(-7\right)\pm\sqrt{\left(-7\right)^2-4.1.16}}{2}\)

\(\Leftrightarrow x=\frac{7\pm\sqrt{-15}}{2}\left(ktm\right)\)

\(\Leftrightarrow x\in\varnothing\)

4 tháng 2 2020

Bài 1:

a) \(7x-5=13-5x\)

\(\Leftrightarrow7x+5x=13+5\)

\(\Leftrightarrow12x=18\)

\(\Leftrightarrow x=18:12\)

\(\Leftrightarrow x=\frac{3}{2}.\)

Vậy phương trình có tập hợp nghiệm là: \(S=\left\{\frac{3}{2}\right\}.\)

b) \(5.\left(2x-3\right)-4.\left(5x-7\right)=19-2.\left(x+11\right)\)

\(\Leftrightarrow10x-15-\left(20x-28\right)=19-\left(2x+22\right)\)

\(\Leftrightarrow10x-15-20x+28=19-2x-22\)

\(\Leftrightarrow13-10x=-3-2x\)

\(\Leftrightarrow13+3=-2x+10x\)

\(\Leftrightarrow16=8x\)

\(\Leftrightarrow x=16:8\)

\(\Leftrightarrow x=2.\)

Vậy phương trình có tập hợp nghiệm là: \(S=\left\{2\right\}.\)

c) \(\frac{2x-1}{3}-\frac{5x+2}{7}=x+13\)

\(\Leftrightarrow\frac{7.\left(2x-1\right)}{7.3}-\frac{3.\left(5x+2\right)}{3.7}=\frac{21.\left(x+13\right)}{21}\)

\(\Leftrightarrow\frac{14x-7}{21}-\frac{15x+6}{21}=\frac{21x+273}{21}\)

\(\Leftrightarrow14x-7-\left(15x+6\right)=21x+273\)

\(\Leftrightarrow14x-7-15x-6=21x+273\)

\(\Leftrightarrow-x-13=21x+273\)

\(\Leftrightarrow-x-21x=273+13\)

\(\Leftrightarrow-22x=286\)

\(\Leftrightarrow x=286:\left(-22\right)\)

\(\Leftrightarrow x=-13.\)

Vậy phương trình có tập hợp nghiệm là: \(S=\left\{-13\right\}.\)

Chúc bạn học tốt!

19 tháng 3 2018

\(a,x-\frac{x+1}{3}=\frac{2x+1}{5}\)

\(\Leftrightarrow\frac{15x}{15}-\frac{5\left(x+1\right)}{15}=\frac{3\left(2x+1\right)}{15}\)

\(\Leftrightarrow15x-5x-5=6x+3\)

\(\Leftrightarrow15x-5x-6x=3+5\)

\(\Leftrightarrow4x=8\)

\(\Leftrightarrow x=2\)

Vậy pt có No là x = 2

b,\(\frac{2x-1}{3}-\frac{5x+2}{7}=x+13\)

\(\Leftrightarrow\frac{7\left(2x-1\right)}{21}-\frac{3\left(5x+2\right)}{21}=\frac{21\left(x+3\right)}{21}\)

\(\Leftrightarrow14x-7-15x-6=21x+273\)

\(\Leftrightarrow14x-5x-21x=273+7+6\)

\(\Leftrightarrow-22x=286\)

\(\Leftrightarrow x=-13\)

Vậy pt có No là x= -13