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a, \(\left(x-4\right)\left(x-5\right)\left(x-6\right)\left(x-7\right)=1680\)
\(\Leftrightarrow\left[\left(x-4\right)\left(x-7\right)\right]\left[\left(x-5\right)\left(x-6\right)\right]=1680\)
\(\Leftrightarrow\left(x^2-11x+28\right)\left(x^2-11x+30\right)=1680\)
Gọi \(k=x^2-11x+29\)
\(\Rightarrow\left(k-1\right)\left(k+1\right)=1680\)
\(\Rightarrow k^2-1=1680\Rightarrow k^2=1681\)
\(\Rightarrow k=\sqrt{1681}=\pm41\)
* TH1: k = -41
\(\Leftrightarrow x^2-11x+29=-41\)
\(\Leftrightarrow x^2-11x+70=0\)
\(\Leftrightarrow x^2-2.\dfrac{11}{2}x+\dfrac{121}{4}-\dfrac{121}{4}+70=0\)
\(\Leftrightarrow\left(x-\dfrac{11}{2}\right)^2+\dfrac{159}{4}=0\Leftrightarrow\left(x-\dfrac{11}{2}\right)^2=\dfrac{-159}{4}\left(vôli\right)\)
Vì \(\left(x-\dfrac{11}{2}\right)^2\ge0\forall x\) mà \(\dfrac{-159}{4}< 0\Rightarrow\left(x-\dfrac{11}{2}\right)^2=\dfrac{-159}{4}\left(loại\right)\)
* TH2: k = 41
\(\Leftrightarrow x^2-11x+29=41\)
\(\Leftrightarrow x^2-11x-12=0\)
\(\Leftrightarrow x^2+x-12x-12=0\)
\(\Leftrightarrow x\left(x+1\right)-12\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-12\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-12=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-1\end{matrix}\right.\)
\(\Rightarrow\left\{x_1=-1;x_2=12\right\}\)
b, \(\left(x+2\right)\left(x+3\right)\left(x-5\right)\left(x-6\right)=180\)
\(\Leftrightarrow\left[\left(x+2\right)\left(x-5\right)\right]\left[\left(x+3\right)\left(x-6\right)\right]=180\)
\(\Leftrightarrow\left(x^2-3x-10\right)\left(x^2-3x-18\right)=180\)
Đặt \(k=x^2-3x-14\)
Ta có pt: \(\left(k-4\right)\left(k+4\right)=180\)
\(\Leftrightarrow k^2-16=180\Leftrightarrow k^2=196\)
\(\Leftrightarrow k=\sqrt{196}=\pm14\)
* TH1: \(t=14\Leftrightarrow x^2-3x-14=14\)
\(\Leftrightarrow x^2-3x-28=0\)
\(\Leftrightarrow\left(x+4\right)\left(x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=7\end{matrix}\right.\)
* TH2: \(t=-14\Leftrightarrow x^2-3x-14=-14\)
\(\Leftrightarrow x^2-3x=0\Leftrightarrow x\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
\(\Rightarrow\left\{x_1=-4;x_2=7;x_3=0;x_4=3\right\}\)
\(\left(x-3\right)^3-2\left(x-1\right)=x\left(x-2\right)^2-5x^2\)
\(\Leftrightarrow x^3-9x^2+27x-27-2x+2=x^3-4x^2+4x-5x^2\)
\(\Leftrightarrow27x-2x-4x-27+2=0\)
\(\Leftrightarrow21x=25\)
\(\Leftrightarrow x=\frac{25}{21}\)
Hết ý tưởng,phá tung ra,sai chỗ nào tự sửa nhé !
\(\frac{\left(x+1\right)^2}{3}+\frac{\left(x+2\right)\left(x-3\right)}{2}=\frac{\left(5x-1\right)\left(x-4\right)}{6}+\frac{28}{3}\)
\(\Leftrightarrow\frac{2\left(x+1\right)^2+3\left(x+2\right)\left(x-3\right)-\left(5x-1\right)\left(x-4\right)}{6}=\frac{28}{3}\)
\(\Leftrightarrow\frac{2x^2+4x+2+3x^2-3x-18-5x^2-21x+4}{6}=\frac{28}{3}\)
\(\Leftrightarrow\frac{\left(4x-3x-21x\right)+\left(2-18+4\right)}{6}=\frac{56}{6}\)
\(\Leftrightarrow-20x-12=56\)
\(\Leftrightarrow-20x=68\)
\(\Leftrightarrow x=-\frac{17}{5}\)
Tự check lại nhá
a/ Đặt (x^2 - 5x) = a thì ta có
a^2 + 10a + 24 = 0
<=> (a + 4)(a + 6) = 0
Làm nốt
b/ (x - 4)(x - 5)(x - 6)(x - 7) = 1680
<=> (x - 4)(x - 7)(x - 5)(x - 6) = 1680
<=> (x^2 - 11x + 28)(x^2 - 11x + 30) = 1680
Đặt x^2 - 11x + 28 = a thì ta có
a(a + 2) = 1680
<=> (a - 40)(a + 42) = 0
Làm nốt
a) \(A=\dfrac{\left(-2\right)^5}{\left(-2\right)^3}=\left(-2\right)^{5-3}=\left(-2\right)^2=4\)
b) \(y\ne0:B=\dfrac{\left(-y\right)^7}{\left(-y\right)^3}=\left(-y\right)^{7-3}=\left(-y\right)^4=y^4\)
c) \(x\ne0:C=\dfrac{\left(x\right)^{12}}{\left(-x\right)^{10}}=\left(x\right)^{12-10}=\left(x\right)^2=x^4\)
d) \(x\ne0:D=\dfrac{2x^6}{\left(2x\right)^3}=\dfrac{2x^6}{8x^3}=\dfrac{1}{4}\left(x\right)^{6-3}=\dfrac{1}{4}\left(x\right)^3\)
e) \(x\ne0:E=\dfrac{\left(-3x\right)^5}{\left(-3x\right)^2}=\left(-3x\right)^{5-2}=\left(-3x\right)^3=-27x^3\)
f) \(x,y\ne0:F=\dfrac{\left(xy^2\right)^4}{\left(xy^2\right)^2}=\left(xy^2\right)^{4-2}=\left(xy^2\right)^2=x^2y^4\)
i) \(x\ne-2:I=\dfrac{\left(x+2\right)^9}{\left(x+2\right)^6}=\left(x+2\right)^{9-6}=\left(x+2\right)^3\)
a) \(\left(2x+3\right)\left(x-4\right)+\left(x+5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x-5x+10=3x^2-12x-5x+20\)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x+10=3x^2-12x+20\)
\(\Leftrightarrow3x^2-7x-2=3x^2-12x+20\)
\(\Leftrightarrow-7x+12x=20+2\)
\(\Leftrightarrow5x=22\)
\(\Rightarrow x=\dfrac{22}{5}\)
tick cho mk nha
b) \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)
\(\Leftrightarrow24x^2+16x-9x-6-4x^2-23x-28=10x^2+3x-1\)
\(\Leftrightarrow20x^2-16x-34-10x^2-3x+1=0\)
\(\Leftrightarrow10x^2-19x-33=0\)
\(\Delta=\left(-19\right)^2-4.10.\left(-33\right)=1320\)
\(x_1=3;x_2=\dfrac{-11}{10}\)
Tick cho mk nha
a) ĐKXĐ: x khác +2
\(\frac{x-2}{2+x}-\frac{3}{x-2}-\frac{2\left(x-11\right)}{x^2-4}\)
<=> \(\frac{x-2}{2+x}-\frac{3}{x-2}=\frac{2\left(x-11\right)}{\left(x-2\right)\left(x+2\right)}\)
<=> (x - 2)^2 - 3(2 + x) = 2(x - 11)
<=> x^2 - 4x + 4 - 6 - 3x = 2x - 22
<=> x^2 - 7x - 2 = 2x - 22
<=> x^2 - 7x - 2 - 2x + 22 = 0
<=> x^2 - 9x + 20 = 0
<=> (x - 4)(x - 5) = 0
<=> x - 4 = 0 hoặc x - 5 = 0
<=> x = 4 hoặc x = 5
làm nốt đi