giúp dùm mình mhes xin cảm ơn
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\(x^2-\sqrt{x^2-5}=7\)
\(\Leftrightarrow\sqrt{x^2-5}=x^2-7\)
\(\Leftrightarrow\left(\sqrt{x^2-5}\right)^2=\left(x^2-7\right)^2\)
\(\Leftrightarrow x^2-5=\left(x^2\right)^2-2.x^2.7+7^2\)
\(\Leftrightarrow x^2-5=x^4-14x^2+49\)
\(\Leftrightarrow-x^4+x^2+14x^2-5-49=0\)
\(\Leftrightarrow-x^4+15x^2-54=0\)
Đặt : \(t=x^2\left(t\ge0\right)\) , ta có :
\(-t^2+15t-54=0\)
\(\left(a=-1;b=15;c=-54\right)\)
\(\Delta=b^2-4ac\)
\(=15^2-4.\left(-1\right).\left(-54\right)\)
\(=225+4.\left(-54\right)\)
\(=225-216\)
\(=9>0\)
\(\sqrt{\Delta}=\sqrt{9}=3\)
\(t_1=\frac{-15+3}{2.\left(-1\right)}=6\) ( nhận )
\(t_2=\frac{-15-3}{2.\left(-1\right)}=9\) ( nhận )
Vs : \(t_1=6\Rightarrow x^2=6\Rightarrow x=\pm\sqrt{6}\)
Vs : \(t_2=9\Rightarrow x^2=9\Rightarrow x=\pm3\)
Vậy phương trình có 4 nghiệm : \(x_1=3;x_2=-3;x_3=6;x_4=-6\)
Cái đề có gì đó sai sai
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1. A
2. C
3. C
4. B
5. C
6. C
7. A
8. B
9. D
10. C
11. A
12. D
13. C
14. A
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1....learned
2....having
3....inviting..
4....to keep.
5....turning
6....which
2.
1. They don't go for a walk because it is raining heavily.
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Answer:
\(23\frac{1}{3}:\left(-\frac{5}{7}\right)-13\frac{1}{3}:\left(-\frac{5}{7}\right)\)
\(=23\frac{1}{3}.\left(-\frac{7}{5}\right)-13\frac{1}{3}.\left(-\frac{7}{5}\right)\)
\(=\left(-\frac{7}{5}\right).\left(23\frac{1}{3}-13\frac{1}{3}\right)\)
\(=\left(-\frac{7}{5}\right).10\)
\(=-14\)
\(25.\left(-\frac{1}{5}\right)+\frac{1}{2}-2.\left(-\frac{1}{2}\right)^2-\sqrt{\frac{1}{4}}\)
\(=-5+\frac{1}{5}-2.\frac{1}{4}-\frac{1}{2}\)
\(=-\frac{25}{4}+\frac{1}{5}-\frac{1}{2}-\frac{1}{2}\)
\(=-\frac{24}{25}-1\)
\(=\frac{-29}{5}\)