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LC
0
CC
0
OK
0
N
28 tháng 8 2017
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LX
28 tháng 8 2017
1 , \(\sqrt{7}\)-\(\sqrt{2}\)> 1
2 , \(^{\left(\sqrt{11}-x\right)^2}\)
DT
0
AH
Akai Haruma
Giáo viên
19 tháng 4 2020
Bài 3:
$f(\sqrt{11})=a(\sqrt{11})^2=11a=-11\Rightarrow a=-1$
Vậy hàm số có dạng $y=-x^2$
Đáp án a.
AH
Akai Haruma
Giáo viên
19 tháng 4 2020
Bài 2:
$f(-47)-f(-31)=365(-47)^2-365.(-31)^2=365.47^2-365.31^2$
$=365(47^2-31^2)>0$ do $47^2>31^2$
$\Rightarrow f(-47)> f(-31)$
Các phương án còn lại thực hiện tương tự ta thấy sai.
Do đó đáp án a là đáp án duy nhất đúng
AH
Akai Haruma
Giáo viên
30 tháng 7 2021
a.
$x^2-11=0$
$\Leftrightarrow x^2=11$
$\Leftrightarrow x=\pm \sqrt{11}$
b. $x^2-12x+52=0$
$\Leftrightarrow (x^2-12x+36)+16=0$
$\Leftrightarrow (x-6)^2=-16< 0$ (vô lý)
Vậy pt vô nghiệm.
c.
$x^2-3x-28=0$
$\Leftrightarrow x^2+4x-7x-28=0$
$\Leftrightarrow x(x+4)-7(x+4)=0$
$\Leftrightarrow (x+4)(x-7)=0$
$\Leftrightarrow x+4=0$ hoặc $x-7=0$
$\Leftrightarrow x=-4$ hoặc $x=7$
AH
Akai Haruma
Giáo viên
30 tháng 7 2021
d.
$x^2-11x+38=0$
$\Leftrightarrow (x^2-11x+5,5^2)+7,75=0$
$\Leftrightarrow (x-5,5)^2=-7,75< 0$ (vô lý)
Vậy pt vô nghiệm
e.
$6x^2+71x+175=0$
$\Leftrightarrow 6x^2+21x+50x+175=0$
$\Leftrightarrow 3x(2x+7)+25(2x+7)=0$
$\Leftrightarrow (3x+25)(2x+7)=0$
$\Leftrightarrow 3x+25=0$ hoặc $2x+7=0$
$\Leftrightarrow x=-\frac{25}{3}$ hoặc $x=-\frac{7}{2}$
NT
2
NV
Nguyễn Việt Lâm
Giáo viên
4 tháng 3 2022
a.
- Với \(y=1\) vế trái hữu tỉ, vế phải vô tỉ (ktm)
- Với \(y\ge4\Rightarrow y!=8k\Rightarrow\left(\sqrt{3}\right)^y=\left(\sqrt{3}\right)^{8k}=81^k\equiv1\left(mod10\right)\)
Mà \(6^x\equiv6\left(mod10\right)\) ; \(11^x\equiv1\left(mod10\right)\Rightarrow10+11^x+6^x\equiv7\left(mod10\right)\)
\(\Rightarrow\) Pt vô nghiệm
- Với \(y=2\Rightarrow\left(\sqrt{3}\right)^y=3\equiv3\left(mod10\right)\) (vô nghiệm do \(VT\equiv7\left(mod10\right)\) theo cmt)
- Với \(y=3\Rightarrow10+11^x+6^x=27\)
\(\Rightarrow11^x+6^x=17\Rightarrow x=1\)
Vậy \(\left(x;y\right)=\left(1;3\right)\)
NV
Nguyễn Việt Lâm
Giáo viên
4 tháng 3 2022
b.
Với \(x\ge4\Rightarrow x!=8k\Rightarrow2^{x!}=2^{8k}=256^k\equiv6\left(mod10\right)\)
Và \(6^y\equiv6\left(mod10\right)\Rightarrow2^{x!}+6^y\equiv12\left(mod10\right)\Rightarrow\) vế trái ko chia hết cho 10 trong khi VP chia hết cho 10 (loại)
Với \(x=1\Rightarrow2+6^y\equiv8\left(mod10\right)\Rightarrow\) vô nghiệm
Với \(x=2\Rightarrow4+6^y=10^y\Rightarrow y=1\)
Với \(x=3\Rightarrow64+6^y=10^y\Rightarrow y=2\)
Vậy \(\left(x;y\right)=\left(2;1\right);\left(3;2\right)\)