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d) \(\left(-45,7\right)+\left[\left(+5,7\right)+\left(+5,75\right)+\left(-0,75\right)\right]\)
\(=\left(-45,7\right)+\left[5,7+5,75-0,75\right]\)
\(=\left(-45,7\right)+5,7+5,75-0,75\)
\(=\left[\left(-45,7+5,7\right)\right]+\left[5,75-0,75\right]\)
\(=-40+5=-35\)
e) \(11,26-5,13:\left(5\frac{5}{18}-1\frac{8}{9}\cdot1,25+1\frac{16}{63}\right)\)
\(=11,26-5,13:\left(\frac{95}{18}-\frac{17}{9}\cdot\frac{5}{4}+\frac{79}{63}\right)\)
\(=11,26-5,13:\left(\frac{95}{18}-\frac{85}{36}+\frac{79}{63}\right)\)
\(=\frac{563}{50}-\frac{513}{100}:\frac{1051}{252}\)
\(=\frac{563}{50}-\frac{513}{100}\cdot\frac{252}{1051}\)
\(=\frac{563}{50}-\frac{129276}{105100}=\frac{21083}{2102}\)
Số lớn quá!
j) \(\sqrt{8^2+6^2}\cdot\sqrt{16}+\frac{1}{2}\cdot\sqrt{\frac{4}{5}}\)
\(=\sqrt{64+36}\cdot\sqrt{16}+\frac{1}{2}\cdot\sqrt{\frac{4}{5}}\)
\(=\sqrt{100}\cdot4+\frac{1}{2}\cdot\frac{2\sqrt{5}}{5}\)
\(=10\cdot4+\frac{\sqrt{5}}{5}=40+\frac{\sqrt{5}}{5}=\frac{200+\sqrt{5}}{5}\)
h) Cái đây mình có làm rồi
1) Ta có: \(\sqrt{x}=1\)
\(\Leftrightarrow\left(\sqrt{x}\right)^2=1^2\)
\(\Leftrightarrow x=1\left(TM\right)\)
Vậy \(x=1\)
2) Ta có: \(\sqrt{x}=3\)
\(\Leftrightarrow\left(\sqrt{x}\right)^2=3^2\)
\(\Leftrightarrow x=9\)
Vậy \(x=9\)
3) Ta có: \(\sqrt{x+1}=11\)
\(\Leftrightarrow\left(\sqrt{x+1}\right)^2=11^2\)
\(\Leftrightarrow x=121\left(TM\right)\)
Vậy \(x=121\)
4) Ta có: \(\sqrt{x-2}=12\)
\(\Leftrightarrow\left(\sqrt{x-2}\right)^2=12^2\)
\(\Leftrightarrow x=144\left(TM\right)\)
Vậy \(x=144\)
5) Ta có: \(\sqrt{2-3x}=9\)
\(\Leftrightarrow\left(\sqrt{2-3x}\right)^2=9^2\)
\(\Leftrightarrow2-3x=81\)
\(\Leftrightarrow-3x=79\)
\(\Leftrightarrow x=-\frac{79}{3}\left(TM\right)\)
Vậy \(x=-\frac{79}{3}\)
6) Ta có: \(\sqrt{3x+4}=5\)
\(\Leftrightarrow\left(\sqrt{3x+4}\right)^2=5^2\)
\(\Leftrightarrow3x+4=25\)
\(\Leftrightarrow3x=21\)
\(\Leftrightarrow x=7\left(TM\right)\)
Vậy \(x=7\)
- Vì đề không cho \(x\in Z\)nên các giá trị ở các phần trên đều thỏa mãn hết các bạn nha ^_^
a.\(2x^2+5x+8+\sqrt{x}=x^2+3x+35+x^2+2x-7\)
\(=2x^2+5x+8+\sqrt{x}=2x^2+5x+28\Leftrightarrow\sqrt{x}=20\Leftrightarrow x=400.\)
b.\(3\sqrt{x}+7x+5=\sqrt{x}+4x-6+3x+18\)
\(=3\sqrt{x}+7x+5=\sqrt{x}+7x+12\Leftrightarrow2\sqrt{x}=7\Leftrightarrow x=\frac{49}{4}.\)
c.\(8\sqrt{x}+2x-9=5x+7+6\sqrt{x}-3x-12.\)
\(=8\sqrt{x}+2x-9=2x+6\sqrt{x}-5\Leftrightarrow2\sqrt{x}=4\Leftrightarrow x=4.\)
d.\(2\sqrt{3x}+11x-18=5x+3+6\sqrt{3x}+6x-21\)
\(=2\sqrt{3x}+11x-18=11x+6\sqrt{3x}-19\Leftrightarrow4\sqrt{3x}=1\)
\(\Leftrightarrow\sqrt{3x}=\frac{1}{4}\Leftrightarrow3x=\frac{1}{16}\Leftrightarrow x=\frac{1}{48}.\)
a) \(2x^2+5x+8+\sqrt{x}=x^2+3x+35+x^2+2x-7\)
<=> \(2x^2+5x+8+\sqrt{x}=2x^2+5x+28\)
<=> \(2x^2+5x+8+\sqrt{x}-\left(2x^2+5\right)=28\)
<=> \(\sqrt{x}+8=28\)
<=> \(\sqrt{x}=28-8\)
<=> \(\sqrt{x}=20\)
<=> \(\left(\sqrt{x}\right)^2=20^2\)
<=> x = 400
=> x = 400
b) \(3\sqrt{x}+7x+5=\sqrt{x}+4x-6+3x+18\)
<=> \(3\sqrt{x}+7x+5=7x+\sqrt{x}+12\)
<=> \(3\sqrt{x}+5=7x+\sqrt{x}+12-7x\)
<=> \(3\sqrt{x}+5=\sqrt{x}+12\)
<=> \(3\sqrt{x}=\sqrt{x}+12-5\)
<=> \(3\sqrt{x}=\sqrt{x}+7\)
<=> \(3\sqrt{x}-\sqrt{x}=7\)
<=> \(2\sqrt{x}=7\)
<=> \(\sqrt{x}=\frac{7}{2}\)
<=> \(\left(\sqrt{x}\right)^2=\left(\frac{7}{2}\right)^2\)
<=> \(x=\frac{49}{4}\)
=> \(x=\frac{49}{4}\)
c) \(8\sqrt{x}+2x-9=5x+7+6\sqrt{x}-3x-12\)
<=> \(8\sqrt{x}+2x-9=2x+6\sqrt{x}-5\)
<=> \(8\sqrt{x}-9=2x+6\sqrt{x}-5-2x\)
<=> \(8\sqrt{x}-9=6\sqrt{x}-5\)
<=> \(8\sqrt{x}=6\sqrt{x}-5+9\)
<=> \(8\sqrt{x}=6\sqrt{x}+4\)
<=> \(8\sqrt{x}-6\sqrt{x}=4\)
<=> \(2\sqrt{x}=4\)
<=> \(\sqrt{x}=2\)
<=> \(\left(\sqrt{x}\right)^2=2^2\)
<=> x = 4
=> x = 4
d) \(2\sqrt{3x}+11x-18=5x+3+6\sqrt{3x}+6x-21\)
<=> \(2\sqrt{3x}+11x-18=11x+6\sqrt{3x}-18\)
<=> \(2\sqrt{3x}+11x-18-\left(11x-18\right)=6\sqrt{3x}\)
<=>\(2\sqrt{3x}=6\sqrt{3x}\)
<=> \(2\sqrt{3x}-6\sqrt{3x}=0\)
<=>\(-4\sqrt{3x}=0\)
<=> \(\sqrt{3x}=0\)
<=> \(\left(\sqrt{3x}\right)^2=0^2\)
<=> 3x = 0
<=> x = 0
=> x = 0
a) 2|2/3 - x| = 1/2
|2/3 - x| = 1/4
|2/3 - x| = 1/4 hoặc |2/3 - x| = -1/4
Xét 2 TH...
- Ta chứng minh bất đẳng thức phụ dưới đây: \(\frac{1}{\sqrt{x}\left(x+1\right)}=\frac{\sqrt{x}}{x\left(x+1\right)}=\sqrt{x}\left(\frac{1}{x}-\frac{1}{x+1}\right)=\sqrt{x}\left(\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{x+1}}\right)\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{x+1}}\right)\)\(=\left(1+\frac{\sqrt{x}}{\sqrt{x+1}}\right)\left(\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{x+1}}\right)< 2\left(\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{x+1}}\right)\)
Áp dụng : \(\frac{1}{\sqrt{1}.2}< 2.\left(1-\frac{1}{\sqrt{2}}\right)\)
\(\frac{1}{\sqrt{2}.3}< 2.\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\right)\)
...................................
\(\frac{1}{\sqrt{2015}.2016}< 2.\left(\frac{1}{\sqrt{2015}}-\frac{1}{\sqrt{2016}}\right)\)
Cộng các BĐT trên với nhau được : \(\frac{1}{2}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{2016\sqrt{2015}}< 2\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2015}}-\frac{1}{\sqrt{2016}}\right)=2\left(1-\frac{1}{\sqrt{2016}}\right)< 2\left(1-\frac{1}{\sqrt{2025}}\right)=\frac{88}{45}\)
Từ đó suy ra đpcm
Cái ............... là gì vậy bn
