7x2+(2+10)=?
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a: Ta có: \(\left(2x-3\right)^2+6\left(2x-1\right)=7\)
\(\Leftrightarrow\left(2x-3\right)^2+6\left(2x-1\right)-7=0\)
\(\Leftrightarrow4x^2-12x+9+12x-6-7=0\)
\(\Leftrightarrow4x^2=4\)
\(\Leftrightarrow x^2=1\)
hay \(x\in\left\{1;-1\right\}\)
b: Ta có: \(x^2-7x+10=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=2\end{matrix}\right.\)
a) \(\left(2x-3\right)^2+6\left(2x-1\right)=7\\ \Rightarrow4x^2-12x+9+12x-6-7=0\\ \Rightarrow4x^2-4=0\\ \Rightarrow x^2-1=0\\ \Rightarrow x^2=1\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)
b) \(x^2-7x+10=0\\ \Rightarrow\left(x^2-2x\right)-\left(5x-10\right)=0\\ \Rightarrow\left(x-2\right)\left(x-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\x-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)
c) \(-6x^2+13x-5=0\\ \Rightarrow-\left(6x^2-13x+5\right)=0\\ \Rightarrow-\left[\left(6x^2-10x\right)-\left(3x-5\right)\right]=0\\ \Rightarrow-\left[2x\left(3x-5\right)-\left(3x-5\right)\right]=0\\ \Rightarrow-\left(2x-1\right)\left(3x-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}-\left(2x-1\right)=0\\3x-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x-1=0\\3x-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{5}{3}\end{matrix}\right.\)
\(D=\frac{4^7.2^8}{3.2^{15}.16^2+5.2.\left(2^{10}\right)^2}\)
\(=\frac{2^{14}.2^8}{3.2^{15}.2^8+5.2.2^{20}}\)
\(=\frac{2^{22}}{3.2^{23}+5.2^{21}}\)
\(=\frac{2^{22}}{2^{21}.\left(2^2.3+5\right)}\)
\(=\frac{2}{2^2.3+5}\)
\(=\frac{2}{17}\)
\(D=\frac{\left(2^2\right)^7.2^8}{3.2^{15}.\left(2^4\right)^2+5.2^2.2^{20}}\)
\(=\frac{2^{14}.2^8}{3.2^{15}.2^8+5.2^{22}}\)
\(=\frac{2^{22}}{3.2^{23}+5.2^{22}}\)
\(=\frac{2^{22}}{2^{22}\left(3.2+5\right)}=\frac{1}{11}\)
a) P(x) = 7x2 . (x2 – 5x + 2 ) – 5x .(x3 – 7x2 + 3x)
= 7x2 . x2 + 7x2 . (-5x) + 7x2 . 2 – [5x. x3 + 5x . (-7x2) + 5x . 3x]
= 7. (x2 . x2) + [7.(-5)] . (x2 . x) + (7.2).x2 – {5. (x.x3) + [5.(-7)]. (x.x2) + (5.3).(x.x)}
= 7x4 + (-35). x3 + 14x2 – [ 5x4 + (-35)x3 + 15x2 ]
= 7x4 + (-35). x3 + 14x2 - 5x4 + 35x3 - 15x2
= (7x4 – 5x4) + [(-35). x3 + 35x3 ] + (14x2 - 15x2 )
= 2x4 + 0 - x2
= 2x4 – x2
b) Thay x = \( - \dfrac{1}{2}\) vào P(x), ta được:
P(\( - \dfrac{1}{2}\)) = 2. (\( - \dfrac{1}{2}\))4 – (\( - \dfrac{1}{2}\))2 \))
\(\begin{array}{l} = 2.\dfrac{1}{{16}} - \dfrac{1}{4} \\ = \dfrac{1}{8} - \dfrac{{2}}{8} \\ = \dfrac{-1}{8} \end{array}\)
a) Đặt \(x^2=a\left(a\ge0\right)\)
Ta có: \(2x^4-7x^2+4=0\)
Suy ra: \(2a^2-7a+4=0\)
\(\Delta=49-4\cdot2\cdot4=49-32=17\)
Vì \(\Delta>0\) nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}a_1=\dfrac{7-\sqrt{17}}{4}\left(nhận\right)\\a_2=\dfrac{-7+\sqrt{17}}{4}\left(loại\right)\end{matrix}\right.\)
Suy ra: \(x^2=\dfrac{7-\sqrt{17}}{4}\)
\(\Leftrightarrow x=\pm\dfrac{\sqrt{7-\sqrt{17}}}{2}\)
Vậy: \(S=\left\{\dfrac{\sqrt{7-\sqrt{17}}}{2};-\dfrac{\sqrt{7-\sqrt{17}}}{2}\right\}\)
a) (137x25-25x37):10
=25x(137-37):10
=25x100:10
=2500:10
=250
b) 2/7x21/4-2/7:13/4
=2/7x21/4-2/7x4/13
=2/7x(21/4-4/13)
=2/7x257/52
=257/182
c)(124x237+152):(870+235x122)
=(29388+152):(870+28670)
=29540:29540
=1
a) (137 x 25 - 25 x 37 ) : 10 = (25 x (137 - 37) ) : 10
= (25 x 100) : 10
= 2500 : 10
= 250
a)=(3425-925):10
=25200:10
=250
b)=3/2-8/91
=257/182
c)=(29388+152):(870+28670)
=29540:29540
=1
d)=1161+1539/72
=2700/72=75/2
e)=10/77+45/77x9/7
=10/77+405/539
=475/539
li ke cho mình nha Lam Le thi
=14+12
=26
Làm gộp nhé