cho 2 số a ,b thỏa mãn : 3b2-4a2=11ab
tính P=\(\frac{14a+10b}{29a-4b}\)
b) giải phương trình : \(\left(8x+3\right)\left(\frac{5x+7}{2-7x}+1\right)=\left(x+5\right)\left(\frac{5x+7}{2-7x}+1\right)\)
Giải các phương trình sau:
a)\(\frac{\left(9x-0.7\right)}{4}-\frac{\left(5x-1.5\right)}{7}=\frac{\left(7x-1.1\right)}{3}-\frac{5\left(0.4-2x\right)}{6}\)
b)\(\frac{3x-1}{x-1}-\frac{2x+5}{x+3}=1-\frac{4}{\left(x-1\right)\left(x+3\right)}\)
c)\(\frac{3}{4\left(x-5\right)}+\frac{15}{50-2x^2}=-\frac{7}{6\left(x+5\right)}\)
d)\(\frac{8x^2}{3\left(1-4x\right)^2}=\frac{2x}{6x-3}-\frac{1+8x}{4+8x}\)
Bài 3: Giải các phương trình sau bằng cách đưa về dạng ax+b =0 :
a) \(\frac{3x-11}{11}-\frac{x}{3}=\frac{3x-5}{7}-\frac{5x-3}{9}\)
b) \(\frac{9x-0,7}{4}-\frac{5x-1,5}{7}=\frac{7x-1,1}{6}-\frac{5\left(0,4-2x\right)}{5}\)
c) \(\frac{5\left(x-1\right)+2}{6}-\frac{7x-1}{4}=\frac{2\left(2x-1\right)}{7}-5\)
d) \(14\frac{1}{2}-\frac{2\left(x+3\right)}{5}=\frac{3x}{2}-\frac{2\left(x-7\right)}{3}\)
a) Ta có: \(\frac{3x-11}{11}-\frac{x}{3}=\frac{3x-5}{7}-\frac{5x-3}{9}\)
\(\Leftrightarrow\frac{63\left(3x-11\right)}{693}-\frac{231x}{693}-\frac{99\left(3x-5\right)}{693}+\frac{77\left(5x-3\right)}{693}=0\)
\(\Leftrightarrow189x-693-231x-297x+495+385x-231=0\)
\(\Leftrightarrow46x-429=0\)
\(\Leftrightarrow46x=429\)
hay \(x=\frac{429}{46}\)
Vậy: \(x=\frac{429}{46}\)
b) Ta có: \(\frac{9x-0,7}{4}-\frac{5x-1,5}{7}=\frac{7x-1,1}{6}-\frac{5\left(0,4-2x\right)}{5}\)
\(\Leftrightarrow\frac{9x-0,7}{4}-\frac{5x-1,5}{7}-\frac{7x-1,1}{6}+\frac{5\left(0,4-2x\right)}{5}=0\)
\(\Leftrightarrow105\left(9x-0,7\right)-60\left(5x-1,5\right)-70\left(7x-1,1\right)+420\left(0,4-2x\right)=0\)
\(\Leftrightarrow945x-\frac{147}{2}-300x+90-490x+77+168-840x=0\)
\(\Leftrightarrow-685x+261.5=0\)
\(\Leftrightarrow-685x=-261.5\)
hay \(x=\frac{523}{1370}\)
Vậy: \(x=\frac{523}{1370}\)
c) Ta có: \(\frac{5\left(x-1\right)+2}{6}-\frac{7x-1}{4}=\frac{2\left(2x-1\right)}{7}-5\)
\(\Leftrightarrow\frac{14\left(5x-3\right)}{84}-\frac{21\left(7x-1\right)}{84}-\frac{24\left(2x-1\right)}{84}+\frac{420}{84}=0\)
\(\Leftrightarrow70x-42-147x+21-48x+24+420=0\)
\(\Leftrightarrow-125x+423=0\)
\(\Leftrightarrow-125x=-423\)
hay \(x=\frac{423}{125}\)
Vậy: \(x=\frac{423}{125}\)
d) Ta có: \(14\frac{1}{2}-\frac{2\left(x+3\right)}{5}=\frac{3x}{2}-\frac{2\left(x-7\right)}{3}\)
\(\Leftrightarrow\frac{435}{30}-\frac{12\left(x+3\right)}{30}-\frac{45x}{30}+\frac{20\left(x-7\right)}{30}=0\)
\(\Leftrightarrow435-12x-36-45x+20x-140=0\)
\(\Leftrightarrow-37x+259=0\)
\(\Leftrightarrow-37x=-259\)
hay \(x=7\)
Vậy: x=7
Giải phương trình:
a) \(\frac{x+\frac{2\left(3-x\right)}{5}}{14}-\frac{5x-4\left(x-1\right)}{24}=\frac{7x+2+\frac{9-3x}{5}}{12}+\frac{2}{3}\)
b) \(\frac{3}{10}\left(1,2-x\right)-\frac{5+7x}{4}=\frac{1}{20}\left(9x+0,2\right)-\frac{12,5x+4,5}{5}\)
a) \(\frac{x+\frac{2\left(3-x\right)}{5}}{14}-\frac{5x-4\left(x-1\right)}{24}=\frac{7x+2+\frac{9-3x}{5}}{12}+\frac{2}{3}\)
<=> \(\frac{5x+2\left(3-x\right)}{70}-\frac{5x-4\left(x-1\right)}{24}=\frac{35x+10+9-3x}{60}+\frac{2}{3}\)
<=> \(12\left(5x+6-2x\right)-35\left(5x-4x+4\right)\)
<=> \(14\left(35x+10+9-3x\right)+280.2\) <=> \(12\left(3x+6\right)-35\left(x+4\right)\)
<=> \(14\left(32x+19\right)+560\)
<=> \(36x+72-35x-140=448x+226+560\)
<=> \(-447x=894\)
<=> x = -2
Giải phương trình:
1.\(\frac{x-5}{x-5}+\frac{x-6}{x-5}+\frac{x-7}{x-5}+...+\frac{1}{x-5}=4\left(x\in N\right)\)
2.\(\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+\frac{1}{x^2+7x+12}+...+\frac{1}{x^2+15x+56}=\frac{1}{14}\)
3.\(\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{x\left(x+2\right)}\right)=\frac{31}{16}\left(x\in N\right)\)
4.\(8\left(x^2+\frac{1}{x^2}\right)-34\left(x+\frac{1}{x}\right)+51=0\)
5.\(6x^4-5x^3-38x^2-5x+6=0\)
BT: Giải phương trình
a)\(\frac{5\left(x-1\right)+2}{6}-\frac{7x-1}{4}=\frac{2\left(2x+1\right)}{7}-5\)
b)\(\frac{3x-2}{6}-5=\frac{3-2\left(x+7\right)}{4}\)
1.Cho x, y là các số thực không âm . Tìm Max của \(\frac{\left(x^2-y^2\right)\left(1-x^2y^2\right)}{\left(1+x^2\right)\left(1+y^2\right)}\)
2.cho a,b,c >0 thỏa mãn \(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=2\).CMR \(abc\le\frac{1}{8}\)
3.Giải phương trình : \(x^3-4\sqrt[3]{4x-3}+3=0\)
4.Tìm x,y thỏa mãn \(5x-2\sqrt{x}\left(2+y\right)+y^2+1=0\)
5.Giải phương trình \(\left(2x^3-3x+1\right)\left(2x^2+5x+1\right)=9x^2\)
6.cho các số dương a , b , c thỏa mãn a+b+c = 4. CMR \(\sqrt[4]{a^3}+\sqrt[4]{b^3}+\sqrt[4]{c^3}>2\sqrt{2}\)
7. Tìm Max của S = \(5x^2+9y^2-12xy+24x-48y+2016\)
8. Giải phương trình \(4\sqrt{x+1}=x^2-5x+14\)
7. \(S=9y^2-12\left(x+4\right)y+\left(5x^2+24x+2016\right)\)
\(=9y^2-12\left(x+4\right)y+4\left(x+4\right)^2+\left(x^2+8x+16\right)+1936\)
\(=\left[3y-2\left(x+4\right)\right]^2+\left(x-4\right)^2+1936\ge1936\)
Vậy \(S_{min}=1936\) \(\Leftrightarrow\) \(\hept{\begin{cases}3y-2\left(x+4\right)=0\\x-4=0\end{cases}}\) \(\Leftrightarrow\) \(\hept{\begin{cases}x=4\\y=\frac{16}{3}\end{cases}}\)
8. \(x^2-5x+14-4\sqrt{x+1}=0\) (ĐK: x > = -1).
\(\Leftrightarrow\) \(\left(x+1\right)-4\sqrt{x+1}+4+\left(x^2-6x+9\right)=0\)
\(\Leftrightarrow\) \(\left(\sqrt{x+1}-2\right)^2+\left(x-3\right)^2=0\)
Với mọi x thực ta luôn có: \(\left(\sqrt{x+1}-2\right)^2\ge0\) và \(\left(x-3\right)^2\ge0\)
Suy ra \(\left(\sqrt{x+1}-2\right)^2+\left(x-3\right)^2\ge0\)
Đẳng thức xảy ra \(\Leftrightarrow\) \(\hept{\begin{cases}\left(\sqrt{x+1}-2\right)^2=0\\\left(x-3\right)^2=0\end{cases}}\) \(\Leftrightarrow\) x = 3 (Nhận)
1.Cho x, y là các số thực không âm . Tìm Max của \(\frac{\left(x^2-y^2\right)\left(1-x^2y^2\right)}{\left(1+x^2\right)\left(1+y^2\right)}\)
2.cho a,b,c >0 thỏa mãn \(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=2\).CMR \(abc\le\frac{1}{8}\)
3.Giải phương trình : \(x^3-4\sqrt[3]{4x-3}+3=0\)
4.Tìm x,y thỏa mãn \(5x-2\sqrt{x}\left(2+y\right)+y^2+1=0\)
5.Giải phương trình \(\left(2x^3-3x+1\right)\left(2x^2+5x+1\right)=9x^2\)
6.cho các số dương a , b , c thỏa mãn a+b+c = 4. CMR \(\sqrt[4]{a^3}+\sqrt[4]{b^3}+\sqrt[4]{c^3}>2\sqrt{2}\)
7. Tìm Max của S = \(5x^2+9y^2-12xy+24x-48y+2016\)
8. Giải phương trình \(4\sqrt{x+1}=x^2-5x+14\)
7. \(S=9y^2-12\left(x+4\right)y+\left(5x^2+24x+2016\right)\)
\(=9y^2-12\left(x+4\right)y+4\left(x+4\right)^2+\left(x^2+8x+16\right)+1936\)
\(=\left[3y-2\left(x+4\right)\right]^2+\left(x-4\right)^2+1936\ge1936\)
Vậy \(S_{min}=1936\) \(\Leftrightarrow\) \(\hept{\begin{cases}3y-2\left(x+4\right)=0\\x-4=0\end{cases}}\) \(\Leftrightarrow\) \(\hept{\begin{cases}x=4\\y=\frac{16}{3}\end{cases}}\)
Câu 8 bn tìm cách tách thành
\(\left(\sqrt{x+1}-2\right)^2+\left(x-3\right)^2=0\)
Bài 1: Giải phương trình:
a, \(\frac{5x-1}{3}+\frac{7x-1,1}{3}-\frac{1,5-5x}{7}=\frac{9x-0,7}{4}\)
Bài 2: Giải các phương trình sau bằng cách đưa về phương trình tích:
a, \(3\left(x-1\right)\left(2x-1\right)=5\left(x+8\right)\left(x-1\right)\)
b, \(9x^2-1=\left(3x+1\right)\left(4x+1\right)\)
c, \(\left(x+7\right)\left(3x-1\right)=49-x^2\)
d, \(x^3-5x^2+6x=0\)
e, \(2x^3+3x^2-32x=48\)
Bài 2:
a, \(3\left(x-1\right)\left(2x-1\right)=5\left(x+8\right)\left(x-1\right)\)
\(\Leftrightarrow3\left(x-1\right)\left(2x-1\right)-5\left(x+8\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(6x-3\right)-\left(5x+40\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(6x-3-5x-40\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-43\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-43=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=43\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là \(S=\left\{1;43\right\}\)
b, \(9x^2-1=\left(3x+1\right)\left(4x+1\right)\)
\(\Leftrightarrow9x^2-1-\left(3x+1\right)\left(4x+1\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(3x+1\right)-\left(3x+1\right)\left(4x+1\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(3x-1-4x-1\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(-x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\-x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\frac{1}{3}\\x=-2\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là \(S=\left\{-\frac{1}{3};-2\right\}\)
c, \(\left(x+7\right)\left(3x-1\right)=49-x^2\)
\(\Leftrightarrow\left(x+7\right)\left(3x-1\right)-\left(49-x^2\right)=0\)
\(\Leftrightarrow\left(x+7\right)\left(3x-1\right)-\left(7-x\right)\left(7+x\right)=0\)
\(\Leftrightarrow\left(x+7\right)\left(3x-1-7+x\right)=0\)
\(\Leftrightarrow\left(x+7\right)\left(4x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+7=0\\4x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-7\\x=2\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là \(S=\left\{-7;2\right\}\)
d, \(x^3-5x^2+6x=0\)
\(\Leftrightarrow x\left(x^2-5x+6\right)=0\)
\(\Leftrightarrow x\left[\left(x^2-2x\right)-\left(3x-6\right)\right]=0\)
\(\Leftrightarrow x\left[x\left(x-2\right)-3\left(x-2\right)\right]=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=3\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là \(S=\left\{0;2;3\right\}\)
e, \(2x^3+3x^2-32x=48\)
\(\Leftrightarrow2x^3+3x^2-32x-48=0\)
\(\Leftrightarrow\left(2x^3-8x^2\right)+\left(11x^2-44x\right)+\left(12x-48\right)=0\)
\(\Leftrightarrow2x^2\left(x-4\right)+11x\left(x-4\right)+12\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(2x^2+11x+12\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left[\left(2x^2+8x\right)+\left(3x+12\right)\right]=0\)
\(\Leftrightarrow\left(x-4\right)\left[2x\left(x+4\right)+3\left(x+4\right)\right]=0\)
\(\Leftrightarrow\left(x-4\right)\left(x+4\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x+4=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\\x=-\frac{3}{2}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là \(S=\left\{4;-4;3-\frac{3}{2}\right\}\)
Dễ mà bạn
giải phương trình
\(\frac{\left(4x+7\right)^2}{7}-\frac{\left(5x-1\right)^2}{7}=\frac{\left(8x-3\right)\left(3x+4\right)}{56}\)
X\(\Leftrightarrow-96x^2+505x+396=0\)ét vế trái : \(\frac{\left(4x+7\right)^2}{7}-\frac{\left(5x-1\right)^2}{7}=\frac{\left(4x+7-5x+1\right)\left(4x+7+5x-1\right)}{7}=\frac{\left(8-x\right)\left(9x+6\right)}{7}\)
pt <=> \(\frac{\left(8-x\right)\left(9x+6\right)}{7}=\frac{\left(8x-3\right)\left(3x+4\right)}{56}\)
\(\Leftrightarrow24\left(8-x\right)\left(3x+2\right)-\left(8x-3\right)\left(3x+4\right)=0\)
\(\Leftrightarrow-96x^2+505x+396=0\)
Giải phương trình trên được : \(x_1=\frac{505}{192}-\frac{\sqrt{407089}}{192}\); \(x_2=\frac{505}{192}+\frac{\sqrt{407089}}{192}\)
Vậy ..............................