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\(1a,\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)
\(\Leftrightarrow\frac{3\left(2x+1\right)^2}{15}-\frac{5\left(x-1\right)^2}{15}=\frac{7x^2-14x-5}{15}\)
\(\Leftrightarrow\frac{12x^2+12x+3}{15}-\frac{5x^2-10x+5}{15}=\frac{7x^2-14x-5}{15}\)
\(\Leftrightarrow12x^2+12x+3-5x^2+10x-5=7x^2-14x-5\)
\(\Leftrightarrow36x=-3\)
\(x=-\frac{1}{12}\)
Vậy ................
\(b,\frac{7x-1}{6}+2x=\frac{16-x}{5}\)
\(\Leftrightarrow\frac{5\left(7x-1\right)}{30}+\frac{30.2x}{30}=\frac{6\left(16-x\right)}{30}\)
\(\Leftrightarrow35x-5+60x=96-6x\)
\(\Leftrightarrow101x=101\)
\(\Leftrightarrow x=1\)
Vậy ....................
<=> \(\frac{3\left(2x+1\right)^2}{15}-\frac{5\left(x-1\right)^2}{15}-\frac{7x^2-14x-5}{15}=0\)
<=> \(\frac{12x^2+12x+3}{15}-\frac{5x^2-10x+5}{15}-\frac{7x^2-14x-5}{15}=0\)
<=> \(\frac{12x^2+12x+3-5x^2+10x-5-7x^2+14x+5}{15}=0\)
=> 36x + 3 = 0
<=> 36x = -3
<=> x = -1/12
Vậy S = { -1/12 }
\(\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)
\(\Leftrightarrow\frac{4x^2+4x+1}{5}-\frac{x^2-2x+1}{3}=\frac{7x^2-14x-5}{15}\)
\(\Leftrightarrow\frac{12x^2+12x+3}{15}-\frac{5x^2-10x+5}{15}=\frac{7x^2-14x-5}{15}\)
\(\Leftrightarrow12x^2+12x+3-5x^2+10x-5-7x^2+14x+5=0\)
\(\Leftrightarrow36x+3=0\Leftrightarrow x=-\frac{3}{36}=-\frac{1}{12}\)
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 :
\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)
\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)
\(\Leftrightarrow30x+15-20=15x-2\)
\(\Leftrightarrow15x=3\)
\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)
Bài 5 :
a, Ta có : \(\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)
=> \(\frac{3\left(2x+1\right)^2}{15}-\frac{5\left(x-1\right)^2}{15}=\frac{7x^2-14x-5}{15}\)
=> \(3\left(2x+1\right)^2-5\left(x-1\right)^2=7x^2-14x-5\)
=> \(12x^2+12x+3-5x^2+10x-5-7x^2+14x+5=0\)
=> \(36x+3=0\)
=> \(x=-\frac{1}{12}\)
Vậy phương trình trên có nghiệm là \(S=\left\{-\frac{1}{12}\right\}\)
b, Ta có : \(\frac{7x-1}{6}+2x=\frac{16-x}{5}\)
=> \(\frac{5\left(7x-1\right)}{30}+\frac{60x}{30}=\frac{6\left(16-x\right)}{30}\)
=> \(5\left(7x-1\right)+60x=6\left(16-x\right)\)
=> \(35x-5+60x-96+6x=0\)
=> \(101x-101=0\)
=> \(x=1\)
Vậy phương trình trên có tạp nghiệm là \(S=\left\{1\right\}\)
c, Ta có : \(\frac{\left(x-2\right)^2}{3}-\frac{\left(2x-3\right)\left(2x+3\right)}{8}+\frac{\left(x-4\right)^2}{6}=0\)
=> \(\frac{8\left(x-2\right)^2}{24}-\frac{3\left(2x-3\right)\left(2x+3\right)}{24}+\frac{4\left(x-4\right)^2}{24}=0\)
=> \(8\left(x-2\right)^2-3\left(2x-3\right)\left(2x+3\right)+4\left(x-4\right)^2=0\)
=> \(8\left(x^2-4x+4\right)-3\left(4x^2-9\right)+4\left(x^2-8x+16\right)=0\)
=> \(8x^2-32x+32-12x^2+27+4x^2-32x+64=0\)
=> \(-64x+123=0\)
=> \(x=\frac{123}{64}\)
Vậy phương trình có nghiệm là \(S=\left\{\frac{123}{64}\right\}\)
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 \(S=\left\{\frac{479}{414}\right\}\)
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