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MT
16 tháng 10 2015
\(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=1+\frac{x}{y}+\frac{x}{z}+\frac{y}{x}+1+\frac{y}{z}+\frac{z}{x}+\frac{z}{y}+1\)
\(=3+\left(\frac{x}{y}+\frac{y}{x}\right)+\left(\frac{x}{z}+\frac{z}{x}\right)+\left(\frac{y}{z}+\frac{z}{y}\right)\)
Áp dụng BĐT cô-si cho hai số không âm ta có:
\(\frac{x}{y}+\frac{y}{x}\ge2\sqrt{\frac{x}{y}.\frac{y}{x}}=2\sqrt{1}=2\)
\(\frac{x}{z}+\frac{z}{x}\ge2\sqrt{\frac{x}{z}.\frac{z}{x}}=2\sqrt{1}=2\)
\(\frac{y}{z}+\frac{z}{y}\ge2\sqrt{\frac{y}{z}.\frac{z}{y}}=2\sqrt{1}=2\)
Suy ra: \(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge3+2+2+2=9\)
=>Điều phải chứng minh
16 tháng 10 2015
đặt A= vế trái
nhân phá ngoặc A ta đc:
A=1+x/y+x/z+y/x+1+y/z+z/x+z/y+1
=3+(x/y+y/x)+(x/z+z/x)+(y/z+z/y)
áp dụng BĐT:a/b+b/a>=2
=>A>=3+2+2+2=9
vậy...
DL
26 tháng 3 2017
\(\frac{3-x+x}{3-x}=\frac{5x\left(x+2\right)+2\left(x+2\right)\left(3-x\right)}{\left(x+2\right)^2\left(3-x\right)}\)
\(\frac{3}{3-x}=\frac{\left(5x+2\left(3-x\right)\right)\left(x+2\right)}{\left(x+2\right)^2\left(3-x\right)}\)
\(\frac{3}{3-x}=\frac{5x+2\left(3-x\right)}{\left(x+2\right)\left(3-x\right)}\)
\(\frac{3}{3-x}=\frac{5x}{\left(x+2\right)\left(3-x\right)}+2\)
\(\frac{3}{3-x}-2=\frac{5x}{\left(x+2\right)\left(3-x\right)}\)
\(\frac{3-2\left(3-x\right)}{\left(x+2\right)\left(3-x\right)}=\frac{5x}{\left(x+2\right)\left(3-x\right)}\)
\(3-2X\left(3-x\right)=5x\)
\(3-6+2x=5x\)
chị có thể tự giải tiếp ạ
e là hs lớp 7
6 tháng 4 2017
cảm ơn e "dang long vu'' chị làm xong thấy cái j nó sai sai nhưng k biết sai chỗ nào nên muốn dò lại bài thôi cảm ơn e nha
10 tháng 3 2020
\(x^2+\frac{9x^3}{\left(x+3\right)^2}=40\left(x\ne-3\right)\)
\(\Leftrightarrow x^2+\left(x+3\right)^2+9x^2=40\left(x+3\right)^2\)
\(\Leftrightarrow x^4+6x^3+18x^2=40x^2+240x+360\)
\(\Leftrightarrow x^4+6x^3-22x^2-240x-360=0\)
\(\Leftrightarrow\left(x^3+10x+30\right)\left(x-6\right)\left(x+2\right)=0\)
Khi x-6=0 hoặc x+2=0 <=> x=6 hoặc x=-2
Khi \(x^3+10x+30=0\)
\(x=\frac{-10+2\sqrt{5}}{2};x=\frac{-10-2\sqrt{5}}{2}\)
Hơi khó hiểu 1 chút, bạn cố gắng nhé
10 tháng 3 2020
\(x^2+\frac{9x^2}{\left(x+3\right)^2}=40^{\left(1\right)}\)
\(ĐKXĐ:x\ne-3\)
\(\left(1\right)\Leftrightarrow x^2-2.x.\frac{3x}{x+3}+\frac{\left(3x\right)^2}{\left(x+3\right)^2}+\frac{6x^2}{x+3}=40\)
\(\Leftrightarrow\left(x-\frac{3x}{x+3}\right)^2+\frac{6x^2}{x+3}=40\)
\(\Leftrightarrow\left(\frac{x^2}{x+3}\right)^2+6.\frac{x^2}{x+3}=40\)
Đặt \(t=\frac{x^2}{x+3}\)ta có
\(t^2+6t=40\)
\(\Leftrightarrow\left(t-4\right)\left(t+10\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}t-4=0\\t+10=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}t=4\\t=-10\end{cases}}\)
+) Với t =4 ta có
\(\frac{x^2}{x+3}=4\)
\(\Rightarrow4\left(x+3\right)=x^2\)
\(\Leftrightarrow x^2-4x-12=0\)
\(\Leftrightarrow\left(x-6\right)\left(x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-6=0\\x+2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=6\left(tm\right)\\x=-2\left(tm\right)\end{cases}}\)
+) với x=-10 ta có
\(\frac{x^2}{x+3}=-10\)
\(\Rightarrow-10\left(x+3\right)=x^2\)
\(\Leftrightarrow x^2+10x+30=0\)
\(\Leftrightarrow\left(x+5\right)^2=-5\)
Phương trình vô nghiệm
Vậy............................
KN
18 tháng 8 2020
1. \(\frac{7x-1}{6}+2x=\frac{16-x}{5}\)
\(\Leftrightarrow5\left(7x-1\right)+60x=6\left(16-x\right)\)
\(\Leftrightarrow35x-5+60x=96-6x\)
\(\Leftrightarrow95x-5=96-6x\)
\(\Leftrightarrow95x+6x=96+5\)
\(\Leftrightarrow101x=101\)
\(\Leftrightarrow x=1\)
2. \(\frac{10x+3}{12}=1+\frac{6+8x}{9}\)
\(\Leftrightarrow3\left(10x+3\right)=36+4\left(6+8x\right)\)
\(\Leftrightarrow30x+9=36+24+32x\)
\(\Leftrightarrow30x+9=32x+60\)
\(\Leftrightarrow30x-32x=60-9\)
\(\Leftrightarrow-2x=51\)
\(\Leftrightarrow x=-\frac{51}{2}\)
3. \(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)
\(\Leftrightarrow8x-3-2\left(3x-2\right)=2\left(2x-1\right)+x+3\)
\(\Leftrightarrow8x-3-6x+4=4x-2+x+3\)
\(\Leftrightarrow2x+1=5x+1\)
\(\Leftrightarrow2x=5x\)
\(\Leftrightarrow x=0\)
19 tháng 8 2020
4) \(\frac{3\left(3-x\right)}{8}+\frac{2\left(5-x\right)}{3}=\frac{1-x}{2}-2\)
=> \(\frac{9-3x}{8}+\frac{10-2x}{3}=\frac{1-x}{2}-\frac{2}{1}\)
=> \(\frac{3\left(9-3x\right)}{24}+\frac{8\left(10-2x\right)}{24}=\frac{12\left(1-x\right)}{24}-\frac{48}{24}\)
=> \(\frac{27-9x}{24}+\frac{80-16x}{24}=\frac{12-12x}{24}-\frac{48}{24}\)
=> \(\frac{27-9x+80-16x}{24}=\frac{12-12x-48}{24}\)
=> 27 - 9x + 80 - 16x = 12 - 12x - 48
=> 27 - 9x + 80 - 16x - 12 + 12x + 48 = 0
=> (27 + 80 - 12 + 48) + (-9x - 16x + 12x) = 0
=> 143 - 13x = 0
=> 13x = 143
=> x = 11
5) \(\frac{2\left(x-3\right)}{7}+\frac{x-5}{3}-\frac{13x+4}{21}=0\)
=> \(\frac{2x-6}{7}+\frac{x-5}{3}-\frac{13x+4}{21}=0\)
=> \(\frac{3\left(2x-6\right)}{21}+\frac{7\left(x-5\right)}{21}-\frac{13x+4}{21}=0\)
=> \(\frac{6x-18}{21}+\frac{7x-35}{21}-\frac{13x+4}{21}=0\)
=> \(\frac{6x-18+7x-35-13x-4}{21}=0\)
=> 6x - 18 + 7x - 35 - 13x - 4 = 0
=> (6x + 7x - 13x) + (-18 - 35 - 4) = 0
=> -57 = 0(vô nghiệm)
6) \(\frac{6x+5}{2}-\left(2x+\frac{2x+1}{2}\right)=\frac{10x+3}{4}\)
=> \(\frac{6x+5}{2}-\frac{10x+3}{4}=2x+\frac{2x+1}{2}\)
=> \(\frac{2\left(6x+5\right)}{4}-\frac{10x+3}{4}=\frac{8x}{4}+\frac{2\left(2x+1\right)}{4}\)
=> \(\frac{12x+10}{4}-\frac{10x+3}{4}=\frac{8x}{4}+\frac{4x+2}{4}\)
=> \(\frac{12x+10-\left(10x+3\right)}{4}=\frac{8x+4x+2}{4}\)
=> \(\frac{12x+10-10x-3}{4}=\frac{12x+2}{4}\)
=> \(12x+10-10x-3=12x+2\)
=> \(2x+10-3=12x+2\)
=> 2x + 10 - 3 - 12x - 2 = 0
=> (2x - 12x) + (10 - 3 - 2) = 0
=> -10x + 5 = 0
=> -10x = -5
=> x = 1/2
7) \(\frac{2x-1}{5}-\frac{x-2}{3}-\frac{x+7}{15}=0\)
=> \(\frac{3\left(2x-1\right)}{15}-\frac{5\left(x-2\right)}{15}-\frac{x+7}{15}=0\)
=> \(\frac{6x-3}{15}-\frac{5x-10}{15}-\frac{x+7}{15}=0\)
=> \(\frac{6x-3-\left(5x-10\right)-\left(x+7\right)}{15}=0\)
=> 6x - 3 - 5x + 10 - x - 7 = 0
=> (6x - 5x - x) + (-3 + 10 - 7) = 0
=> 0x + 0 = 0
=> 0x = 0
=> x tùy ý
Bài 8 tự làm nhé
VN
16 tháng 8 2016
a) \(\frac{5-x}{4x^2-8x}\) + \(\frac{7}{8x}\) = \(\frac{x-1}{2x\left(x-2\right)}\) +\(\frac{1}{8x-16}\) ĐKXĐ : x #0, x#2, x#-2
<=> \(\frac{5-x}{4x\left(x-2\right)}\) + \(\frac{7}{8x}=\frac{x-1}{2x\left(x-2\right)}\) + \(\frac{1}{8\left(x-2\right)}\)
<=> \(\frac{2\left(5-x\right)}{8x\left(x-2\right)}+\frac{7\left(x-2\right)}{8x\left(x-2\right)}=\frac{4\left(x-1\right)}{8x\left(x-2\right)}+\frac{x}{8x\left(x-2\right)}\)
=> 10 - 2x + 7x - 14 = 4x - 4 + x
<=>-2x + 7x - 4x + x = -4 - 10 + 14
<=>x=-14
\(\frac{3000-10x}{x}.\left(x+10\right)+8x=3000\)
\(\Leftrightarrow\left(3000-10x\right).\left(x+10\right)+8x^2=3000x\)
\(\Leftrightarrow3000x+30000-10x^2-100x+8x^2-3000x=0\)
\(\Leftrightarrow-2x^2-100x+30000=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=100\\x=-150\end{cases}}\)
3000x-10x^2+30000-100x 8x^2
<>--------------------------------------- + ------------ = 3000
x x
<> -10x^2+2900x+30000+8x^2=3000x
<> -2x^2-100x+30000=0
<>2x^2+100x-30000=0
<>x=-150
<>x=100