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a: \(\left(x+1\right)^3+\left(x-2\right)^3=2x^3+2\left(2x-1\right)^2-9\)
\(\Leftrightarrow x^3+3x^2+3x+1+x^3-6x^2+12x-8=2x^3+2\left(4x^2-4x+1\right)-9\)
\(\Leftrightarrow2x^3-3x^2+15x-7=2x^3+8x^2-8x-7\)
\(\Leftrightarrow-11x^2+23x=0\)
\(\Leftrightarrow x\left(-11x+23\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{23}{11}\end{matrix}\right.\)
c)\(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(\Rightarrow\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)
\(\Rightarrow2x+\frac{3}{5}=\pm\frac{3}{5}\)
- Với \(2x+\frac{3}{5}=\frac{3}{5}\)
\(\Rightarrow2x=0\Rightarrow x=0\)
- Với \(2x+\frac{3}{5}=-\frac{3}{5}\)
\(\Rightarrow2x=-\frac{6}{5}\Rightarrow x=-\frac{3}{5}\)
a)x=10
b)x=61/114
c)x=0
d)sai cái gì đó
Đáp án là gì nhưng lời giải ???????
a) ( 3x + 2 )( x - 1 ) - ( x + 2 )( 3x + 1 ) = 7
<=> 3x2 - x - 2 - ( 3x2 + 7x + 2 ) = 7
<=> 3x2 - x - 2 - 3x2 - 7x - 2 = 7
<=> -8x - 4 = 7
<=> -8x = 11
<=> x = -11/8
b) ( 6x + 5 )( 2x + 3 ) - ( 4x + 3 )( 3x - 2 ) = 8
<=> 12x2 + 28x + 15 - ( 12x2 + x - 6 ) = 8
<=> 12x2 + 28x + 15 - 12x2 - x + 6 = 8
<=> 27x + 21 = 8
<=> 27x = -13
<=> x = -13/27
c) 2x( x + 3 ) - ( x + 1 )( 2x + 1 ) - 5 = 9
<=> 2x2 + 6x - ( 2x2 + 3x + 1 ) - 5 = 9
<=> 2x2 + 6x - 2x2 - 3x - 1 - 5 = 9
<=> 3x - 6 = 9
<=> 3x = 15
<=> x = 5
d) ( 5x + 3 )( 4x - 7 ) - ( 10x + 9 )( 2x - 3 ) = 10
<=> 20x2 - 23x - 21 - ( 20x2 - 12x - 27 ) = 10
<=> 20x2 - 23x - 21 - 20x2 + 12x + 27 = 10
<=> -11x + 6 = 10
<=> -11x = 4
<=> x = -4/11
a, \(\left(3x+2\right)\left(x-1\right)-\left(x+2\right)\left(3x+1\right)=7\Leftrightarrow-8x-4=7\Leftrightarrow x=-\frac{11}{8}\)
b, \(\left(6x+5\right)\left(2x+3\right)-\left(4x+3\right)\left(3x-2\right)=8\Leftrightarrow27x+21=8\Leftrightarrow x=-\frac{13}{27}\)
c, \(2x\left(x+3\right)-\left(x+1\right)\left(2x+1\right)-5=9\Leftrightarrow3x-6=9\Leftrightarrow x=5\)
d, \(\left(5x+3\right)\left(4x-7\right)-\left(10x+9\right)\left(2x-3\right)=10\Leftrightarrow-11x+6=10\Leftrightarrow x=-\frac{4}{11}\)
a: Ta có: \(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=6\)
\(\Leftrightarrow x^2+5x+6-x^2-3x+10=6\)
\(\Leftrightarrow2x=-10\)
hay x=-5
b: Ta có: \(\left(3x+2\right)\left(2x+9\right)-\left(x+2\right)\left(6x+1\right)=\left(x+1\right)-\left(x-6\right)\)
\(\Leftrightarrow6x^2+27x+4x+18-6x^2-x-12x-2=x+1-x+6\)
\(\Leftrightarrow18x+16=7\)
hay \(x=-\dfrac{1}{2}\)
c: Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)
\(\Leftrightarrow3\left(6x^2-2x-3x+1\right)-\left(18x^2-2x-27x+3\right)=0\)
\(\Leftrightarrow18x^2-15x+3-18x^2+27x-3=0\)
hay x=0
a: \(\Leftrightarrow2x-3+14⋮2x-3\)
\(\Leftrightarrow2x-3\in\left\{1;-1;2;-2;7;-7;14;-14\right\}\)
hay \(x\in\left\{2;1;\dfrac{5}{2};\dfrac{1}{2};5;-2;\dfrac{17}{2};-\dfrac{11}{2}\right\}\)
b: \(\Leftrightarrow3x+9⋮2x-1\)
\(\Leftrightarrow6x+18⋮2x-1\)
\(\Leftrightarrow2x-1\in\left\{1;-1;3;-3;7;-7;21;-21\right\}\)
hay \(x\in\left\{1;0;2;-1;4;-3;11;-10\right\}\)
c: \(\Leftrightarrow3x+9⋮3x-4\)
\(\Leftrightarrow3x-4\in\left\{1;-1;13;-13\right\}\)
hay \(x\in\left\{\dfrac{5}{3};1;\dfrac{17}{3};-3\right\}\)
không ai trả lời
a,\(2\left(3x-1\right)-5\left(x-3\right)-9\left(2x-4\right)=24\)
\(< =>6x-2-5x+15-18x+36=24\)
\(< =>-29x+49=24< =>29x=25< =>x=\frac{25}{29}\)
b,\(2x^2+4\left(x^2-1\right)=2x\left(3x+1\right)\)
\(< =>2x^2+4x^2-4=6x^2+2x\)
\(< =>2x=-4< =>x=-\frac{4}{2}=-2\)
c, \(2x\left(5-3x\right)+2x\left(3x-5\right)-3\left(x-7\right)=4\)
\(< =>10x-6x^2+6x^2-10x-3x+21=4\)
\(< =>-3x=4-21=-17< =>x=\frac{17}{3}\)
d, \(5x\left(x+1\right)-4x\left(x+2\right)=1-x\)
\(< =>5x^2+5x-4x^2-8x=1-x\)
\(< =>x^2-3x+x-1=0\)
\(< =>x^2-2x-1=0\)
\(< =>\left(x-1\right)^2=2\)
\(< =>\orbr{\begin{cases}x-1=\sqrt{2}\\x-1=-\sqrt{2}\end{cases}}\)
\(< =>\orbr{\begin{cases}x=1+\sqrt{2}\\x=1-\sqrt{2}\end{cases}}\)
:V
\(\Leftrightarrow9x^2-1-1+6x^2-27x+2x-9=0\)
\(\Leftrightarrow15x^2-25x-11=0\)
\(\text{Δ}=\left(-25\right)^2-4\cdot15\cdot\left(-11\right)=1285\)
=> Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{25-\sqrt{1285}}{30}\\x_2=\dfrac{25+\sqrt{1285}}{30}\end{matrix}\right.\)