Giải phương trình:
\(\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)
Giải các phương trình sau bằng cách đưa về dạng ax + b = 0:
\(a,\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)
\(b,\frac{x+1}{3}+\frac{3\left(2x+1\right)}{4}=\frac{2x+3\left(x+1\right)}{6}+\frac{7+12x}{12}\)
a)\(\frac{\left(2x-3\right)\left(2x+3\right)}{8}=\frac{\left(x-4\right)^2}{6}+\frac{\left(x-2\right)^2}{3}\)
b)\(\frac{7x^2-14x-5}{15}=\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}\)
c)\(\frac{\left(7x+1\right)\left(x-2\right)}{10}+\frac{2}{5}=\frac{\left(x-2\right)^2}{5}+\frac{\left(x-1\right)\left(x-3\right)}{2}\)
Giải các phương trình sau :
ĐS: a) x= \(\frac{123}{64}\) b) x=\(\frac{1}{2}\) c) \(\frac{19}{15}\)
Giải các phương trình sau bằng cách đưa về dạng ax + b = 0:
\(a,\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)
\(b,\frac{x+1}{3}+\frac{3\left(2x+1\right)}{4}=\frac{2x+3\left(x+1\right)}{6}+\frac{7+12x}{12}\)
Hãy giải phương trình sau!
a,\(\frac{x+\frac{x+1}{5}}{3}=1-\frac{2x-\frac{1-2x}{34}}{5}\)
b,\(\frac{^{\left(2x+1\right)^2}}{5}-\frac{^{\left(x-1\right)^2}}{3}=\frac{7x^2-14x-5}{15}\)
c,\(\frac{2x-1}{2}+\frac{2x+1}{3}=\frac{2x+7}{6}+\frac{2x+9}{7}\) ( gợi ý : thêm hoặc bớt 2 )
a) \(\frac{x+\frac{x+1}{5}}{3}=1-\frac{2x-\frac{1-2x}{34}}{5}\)
\(\Leftrightarrow\frac{\frac{5x+x+1}{5}}{3}=1-\frac{\frac{68x-1+2x}{34}}{5}\)
\(\Leftrightarrow\frac{6x+1}{15}=1-\frac{70-1}{170}\)
\(\Leftrightarrow\frac{6x+1}{15}+\frac{70x-1}{170}-1=0\)
\(\Leftrightarrow\frac{34\left(6x+1\right)+3\left(70x-1\right)-510}{510}=0\)
\(\Leftrightarrow204x+34+210x-3-510=0\)
\(\Leftrightarrow414x-479=0\)
\(\Leftrightarrow x=\frac{479}{414}\)
Vậy tập nghiệm của phương trình là \(S=\left\{\frac{479}{414}\right\}\)
bạn ơi bạn làm được câu c chưa
b) \(\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)
\(\Leftrightarrow\frac{4x^2+4x+1}{5}-\frac{x^2-2x+1}{3}-\frac{7x^2-14x-5}{15}=0\)
\(\Leftrightarrow\frac{3\left(4x^2+4x+1\right)-5\left(x^2-2x+1\right)-7x^2+14x+5}{15}=0\)
\(\Leftrightarrow12x^2+12x+3-5x^2+10x-5-7x^2+14x+5=0\)
\(\Leftrightarrow36x+3=0\)
\(\Leftrightarrow x=-\frac{1}{12}\)
Vậy tập nghiệm của phương trình là \(S=\left\{-\frac{1}{12}\right\}\)
c) \(\frac{2x-1}{2}+\frac{2x+1}{3}=\frac{2x+7}{6}+\frac{2x+9}{7}\)
\(\Leftrightarrow\left(\frac{2x-1}{2}-2\right)+\left(\frac{2x+1}{3}-2\right)=\left(\frac{2x+7}{6}-2\right)+\left(\frac{2x+9}{7}-2\right)\)
\(\Leftrightarrow\frac{2x-5}{2}+\frac{2x-5}{3}=\frac{2x-5}{6}+\frac{2x-5}{7}\)
\(\Leftrightarrow\frac{2x-5}{2}+\frac{2x-5}{3}-\frac{2x-5}{6}-\frac{2x-5}{7}=0\)
\(\Leftrightarrow\left(2x-5\right)\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{6}-\frac{1}{7}\right)=0\)
Mà \(\frac{1}{2}+\frac{1}{3}-\frac{1}{6}-\frac{1}{7}\ne0\)
\(\Leftrightarrow2x-5=0\)
\(\Leftrightarrow x=\frac{5}{2}\)
Vậy tập nghiệm của phương trình là \(S=\left\{\frac{5}{2}\right\}\)
\(\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}< \frac{7x^2-14x-5}{15}\)
\(\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}< \frac{7x^2-14x-5}{15}\)
\(\Leftrightarrow\frac{3\left(2x+1\right)^2}{15}-\frac{5\left(x-1\right)^2}{15}< \frac{7x^2-14x-5}{15}\)
\(\Rightarrow3\left(2x+1\right)^2-5\left(x-1\right)^2< 7x^2-14x-5\)
\(\Leftrightarrow3\left(4x^2+4x+1\right)-5\left(x^2-2x+1\right)< 7x^2-14x-5\)
\(\Leftrightarrow\left(12x^2+12x+3\right)-\left(5x^2-10x+5\right)< 7x^2-14x-5\)
\(\Leftrightarrow12x^2+12x+3-5x^2+10x-5< 7x^2-14x-5\)
\(\Leftrightarrow7x^2+22x-2< 7x^2-14x-5\)
\(\Leftrightarrow7x^2+22x-2-7x^2+14x+5< 0\)
\(\Leftrightarrow36x+3< 0\)
\(\Leftrightarrow36x< -3\)
\(\Leftrightarrow x< -\frac{3}{36}\)
\(\Leftrightarrow x< -\frac{1}{12}\)
bn ơi kl phải là <=>x >\(-\frac{1}{12}\)
Giải các phương trình sau:
a\(\frac{\left(2x+1\right)^2}{5}\) - \(\frac{\left(x-1\right)^2}{3}\) = \(\frac{7x^2-14x-5}{15}\)
phương trình tương đương
<=>\(\frac{3\left(2x+1\right)^2}{15}-\frac{5\left(x-1\right)^2}{15}=\frac{7x^2-14x-5}{15}\)
<=>\(\frac{3\left(4x^2+4x+1\right)}{15}-\frac{5\left(x^2-2x+1\right)}{15}=\frac{7x^2-14x-5}{15}\)
<=>\(\frac{12x^2+12x+3-5x^2+10x-5}{15}=\frac{7x^2-14x-5}{15}\)
<=>\(\frac{7x^2+22x-2}{15}=\frac{7x^2-14x-5}{15}\)<=>7\(x^2+22x-2=7x^2-14x-5\)
<=>22x+14x=-5+2
<=>36x=-3
<=>>x=\(\frac{-3}{36}=\frac{-1}{12}\)
giải phương trình sau :
a) 5-(x-6) = 4(3-2x) b) 2x(x+2)2-8x2 = 2(x-2)(x2+4)
c) 7-(2x+4) = -(x+4) d) (x+1)(2x-3) = (2x-1)(x+5
f) \(\frac{7x-1}{6}+2x=\frac{16-x}{5}\)
e) \(\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)
Giải phương trình:
1. \(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)
2. \(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)
3. \(\frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{15}\)
4. \(\frac{x+3}{2}-\frac{x-1}{3}=\frac{x+5}{6}+1\)
5. \(\frac{x-4}{5}-\frac{3x-2}{10}-x=\frac{2x-5}{3}-\frac{7x+2}{6}\)
6. \(\frac{\left(x+2\right)\left(x+10\right)}{3}-\frac{\left(x+4\right)\left(x+10\right)}{12}=\frac{\left(x-2\right)\left(x+4\right)}{4}\)
7. \(\frac{\left(x+2\right)^2}{8}-2\left(2x-1\right)=25+\frac{\left(x-2\right)^2}{8}\)
8.\(\frac{7x^2-14x-5}{5}=\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}\)
9. \(\frac{\left(2x-3\right)\left(2x+3\right)}{8}=\frac{\left(x-4\right)^2}{6}+\frac{\left(x-2\right)^2}{3}\)
10. \(\frac{x+1}{35}+\frac{x+3}{33}=\frac{x+5}{31}+\frac{x+7}{29}\)
1.
\(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)
\(MC:12\)
Quy đồng :
\(\Rightarrow\frac{3.\left(2x+3\right)}{12}-\left(\frac{2.\left(5x+3\right)}{12}\right)=\frac{3x-4}{12}\)
\(\frac{6x+9}{12}-\left(\frac{10x+6}{12}\right)=\frac{3x-4}{12}\)
\(\Leftrightarrow6x+9-\left(10x+6\right)=3x-4\)
\(\Leftrightarrow6x+9-3x=-4-9+16\)
\(\Leftrightarrow-7x=3\)
\(\Leftrightarrow x=\frac{-3}{7}\)
2.\(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)
\(MC:20\)
Quy đồng :
\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)
\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)
\(\Leftrightarrow30x+15-20=15x-2\)
\(\Leftrightarrow15x=3\)
\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)
Giải phương trình sau:
\(\dfrac{\left(2x+1\right)^2}{5}-\dfrac{\left(x-1\right)^2}{3}=\dfrac{7x^2-14x-5}{15}\)
\(\dfrac{\left(2x+1\right)^2}{5}-\dfrac{\left(x-1\right)^2}{3}=\dfrac{7x^2-14x-5}{15}\)
⇔ \(\dfrac{3\left(2x+1\right)^2}{15}-\dfrac{5\left(x-1\right)^2}{15}=\dfrac{7x^2-14x-5}{15}\)
⇔ \(3\left(2x+1\right)^2-5\left(x-1\right)^2=7x^2-14x-5\)
⇔ \(3\left(4x^2+4x+1\right)-5\left(x^2-2x+1\right)=7x^2-14x-5\)
⇔ \(12x^2+12x+3-5x^2+10x-5=7x^2-14x-5\)
⇔ \(7x^2+22x-2=7x^2-14x-5\) ⇔ \(36x+3=0\) ⇔ x=\(\dfrac{-1}{12}\)
\(\Leftrightarrow3\left(4x^2+4x+1\right)-5\left(x^2-2x+1\right)=7x^2-14x-5\)
\(\Leftrightarrow12x^2+12x+3-5x^2+10x-5-7x^2+14x+5=0\)
\(\Leftrightarrow36x=-3\)
hay x=-1/12