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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) \(\sqrt{\left(\sqrt{9-1}\right)^2}\) = \(\sqrt{8^2}\) = \(\sqrt{64}\) = 8
b) \(\sqrt{x+3}\) = 5 \(\Rightarrow\) \(x\) + 3 = 52 = 25
\(\Rightarrow\) \(x\) = 25 - 3 = 22
c) \(\sqrt{3x-2}\) - 7 = 0 \(\Rightarrow\) \(\sqrt{3x-2}\) = 0 + 7 = 7
\(\Rightarrow\) 3\(x\) - 2 = 72 = 49 \(\Rightarrow\) 3\(x\) = 49 + 2 = 51
\(\Rightarrow\) \(x\) = \(\frac{51}{3}\) = 17
d) \(\sqrt{2x}\) = 8 \(\Rightarrow\) 2\(x\) = 82 = 64 \(\Rightarrow\) \(x\) = \(\frac{64}{2}\) = 32
a)\(\sqrt{x}=4\Leftrightarrow x=4^2\Leftrightarrow x=16\)
b)\(\sqrt{x-2}=3\Leftrightarrow x-2=3^2\Leftrightarrow x=9-2=7\)
c)\(\sqrt{\dfrac{x}{3}-\dfrac{7}{6}}=\dfrac{1}{6}\Leftrightarrow\dfrac{x}{3}-\dfrac{7}{6}=\dfrac{1}{36}\Leftrightarrow\dfrac{x}{3}=-\dfrac{41}{36}\Leftrightarrow x=-\dfrac{41}{12}\)
d)\(x^2=7vớix< 0\)
\(\Leftrightarrow\left(-x\right)^2=7\Leftrightarrow-x=\sqrt{7}\Leftrightarrow x=-\sqrt{7}\)
e)\(x^2-4=0với>0\)
\(\Leftrightarrow x^2=4\Leftrightarrow x=\sqrt{4}=2\)
f)\(\left(2x+7\sqrt{7}\right)^2=7\)
\(\Leftrightarrow4x^2+\sqrt{5488}+343=7\)
\(\Leftrightarrow4x^2+\sqrt{5488}=-336\)
\(\Leftrightarrow4x^2=28\left(12-\sqrt{7}\right)\Leftrightarrow x^2=\dfrac{28\left(12-\sqrt{7}\right)}{4}=7\left(12-\sqrt{7}\right)\)
\(\Leftrightarrow x=\sqrt{7\left(12-\sqrt{7}\right)}=\sqrt{84-7\sqrt{7}}\)
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...
a) \(2x=\sqrt{x}\left(ĐK:x\ne0\right)\)
\(\Leftrightarrow\left(2x\right)^2=x\)
\(\Leftrightarrow4.x^2=x\)
\(\Leftrightarrow4=x:x^2=x.\frac{1}{x^2}=\frac{x}{x^2}=x^{-1}\) ( vô lí vì \(x^{-1}\le0\) )
Vậy : \(x\in\varnothing\)
b) \(\sqrt{x}-1=2\)
\(\Leftrightarrow\sqrt{x}=2+1=3\)
\(\Leftrightarrow x=3^2=9\)
Vậy : \(x=9\)
c) \(3\sqrt{x}-2=7\)
\(\Leftrightarrow3\sqrt{x}-2=7+2=9\)
\(\Leftrightarrow\sqrt{x}=9:3=3\)
\(\Leftrightarrow x=3^2=9\)
Vậy :\(x=9\)
d) \(\sqrt{x-1}+1=3\)
\(\Leftrightarrow\sqrt{x-1}=3-1=2\)
\(\Leftrightarrow x-1=2^2=4\)
\(\Leftrightarrow x=5\)
Vậy : \(x=5\)
19) \(\sqrt{19-x}=19\)
\(\Rightarrow\sqrt{19-x}=\sqrt{19^2}\)
\(\Rightarrow19-x=19^2\)
\(\Rightarrow19-19^2=x\)
\(\Rightarrow x=19\left(1-19\right)=-19.18=-342\)
21) \(\sqrt{x-1}=\dfrac{1}{3}\)
\(\Rightarrow\sqrt{x-1}=\sqrt{\left(\dfrac{1}{3}\right)^2}\)
\(\Rightarrow x-1=\dfrac{1}{3^2}\)
\(x=\dfrac{1+9}{9}=\dfrac{10}{9}\)
24)\(\sqrt{2x+\dfrac{5}{4}}=\dfrac{3}{2}\)
\(\Rightarrow\sqrt{2x+\dfrac{5}{4}}=\sqrt{\left(\dfrac{3}{2}\right)^2}\)
\(\Rightarrow2x+\dfrac{5}{4}=\left(\dfrac{3}{2}\right)^2=\dfrac{9}{4}\)
\(\Rightarrow2x=\dfrac{9-5}{4}=1\)
\(\Rightarrow x=0,5\)
25) \(\sqrt{\dfrac{x}{3}-\dfrac{7}{6}}=\dfrac{1}{6}\)
\(\Rightarrow\sqrt{\dfrac{2x-7}{6}}=\sqrt{\left(\dfrac{1}{6}\right)^2}\)
\(\Rightarrow\dfrac{2x-7}{6}=\left(\dfrac{1}{6}\right)^2=\dfrac{1}{36}\)
\(\Rightarrow\dfrac{12x-42}{36}=\dfrac{1}{36}\)
\(\Rightarrow12x-42=1\)
\(\Rightarrow12x=43\)
\(\Rightarrow x=\dfrac{43}{12}\)
B1
a. = 7/3. ( 37/5 - 32/5)
= 7/3 . 1
= 7/3
Phần b có gì đó sai sao lại có 3:+
c. = 4 + 6 - 3 + 5
= 12
d. = -5/21 : -19/21 : 4/5
= 25/76
B2
a. 1/4 : x =1/2 - 3/4
x = -1/4
x = 1/4 : -1/4
x = -1
b. 2 . | 2x - 3 | = 4 - (-8)
2 . | 2x - 3| = 12
| 2x - 3 | = 12:2
| 2x - 3 | = 6
| x - 3 | = 6:2
| x - 3 | = 3
=> x - 3 = +- 3
* x - 3 = 3
x = 6
* x - 3 = -3
x = 0
Chúc bạn vui vẻ
\(\sqrt[]{2x-3}=7\left(x\ge\dfrac{3}{2}\right)\)
\(\Rightarrow2x-3=49\Rightarrow2x=52\Rightarrow x=26\)