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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) 35x+8=33
<=> 5x+8=3
<=>5x=3-8=-5
<=>x=-1
b) 28-x=25
<=> 8-x=5
<=> x=8-5=3
\(3^{5x+8}=27\)
\(3^{5x+8}=3^3\)
\(5x+8=3\)
\(5x=-5\)
\(x=-1\)
\(2^{8-x}=32\)
\(2^{8-x}=2^5\)
\(8-x=5\)
\(x=3\)
=.= hok tốt!!
- -6x3 + x2 + 5x - 2 = 0
=> -6x3 - 6x2 + 7x2 + 7x - 2x - 2 = 0
=> -6x2(x+1) + 7x(x+1) - 2(x+1) = 0
=> (x+1)(-6x2+7x-2) = 0
=> (x+1)(x2-\(\frac{7}{6}x+\frac{1}{3}\)) = 0
\(\Rightarrow\left(x+1\right)\left(x-\frac{1}{2}\right)\left(x-\frac{2}{3}\right)=0\)
=> x = -1 hoặc x = 1/2 hoặc x = 2/3
- 3x3 + 19x2 + 4x - 12 = 0
=> 3x3 + 3x2 + 16x2 + 16x - 12x - 12 = 0
=> (x+1)(3x2+16x-12)=0
=> (x+1)\(\left(x^2+\frac{16}{3}x-4\right)=0\)
=> (x+1) \(\left(x-\frac{2}{3}\right)\left(x+6\right)=0\)
=> x = -1 hoặcx = 2/3 hoặc x = -6
- 2x3 - 11x2 + 10x + 8 = 0
=> 2x3 - 4x2 - 7x2 + 14x - 4x + 8 = 0
=> 2x2(x - 2) - 7x(x - 2) - 4(x - 2) = 0
=> (x - 2)(2x2 - 7x - 4)=0
=> (x - 2)(\(x^2-\frac{7}{2}x-2\)) = 0
=> \(\left(x-2\right)\left(x-4\right)\left(x+\frac{1}{2}\right)=0\)
=> x = 2 hoặc x = 4 hoặc x = -1/2
a) \(\left(3x-1\right)^4=\frac{1}{16}\)
\(\Rightarrow\orbr{\begin{cases}3x-1=\frac{1}{2}\\3x-1=-\frac{1}{2}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}3x=\frac{1}{2}+1=\frac{3}{2}\\3x=-\frac{1}{2}+1=\frac{1}{2}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{1}{6}\end{cases}}\)
b) \(\left(5x-2\right)^2=\frac{25}{121}\)
\(\Rightarrow\orbr{\begin{cases}5x-2=\frac{5}{11}\\5x-2=-\frac{5}{11}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}5x=\frac{5}{11}+2=\frac{27}{11}\\5x=-\frac{5}{11}+2=\frac{17}{11}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{27}{55}\\x=\frac{17}{55}\end{cases}}\)
c) \(\left(5x+6\right)^3=\frac{-8}{27}=\left(\frac{-2}{3}\right)^3\)
\(\Rightarrow5x+6=-\frac{2}{3}\)
\(\Rightarrow5x=-\frac{2}{3}-6=-\frac{20}{3}\)
\(\Rightarrow x=-\frac{20}{3}:5=-\frac{4}{3}\)
a) (3x — 1) ^4 = 1/16
=>3x-1=-1/2 hoặc 1/2
- Với 3x-1=-1/2
=>3x=1/2
=>x=1/6
- Với 3x-1=1/2
=>3x=3/2
=>x=1/2
b)(5x — 2 )^2 = 25/121
=>5x-2=-5/11 hoặc 5/11
- Với 5x-2=-5/11
=>5x=17/11
=>x=17/55
- Với 5x-2=5/11
=>5x=27/11
=>x=27/55
c)(5x + 6)^3= -8/27
=>5x+6=-2/3
=>5x=-20/3
=>x=-4/3
a: =>|x-3|=12-5x-8=4-5x
\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{4}{5}\\\left(4-5x\right)^2=\left(x-3\right)^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{4}{5}\\\left(5x-4-x+3\right)\left(5x-4+x-3\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{4}{5}\\\left(4x-1\right)\left(6x-7\right)=0\end{matrix}\right.\Leftrightarrow x=\dfrac{1}{4}\)
b: |2-3x|=|6-x|
=>|3x-2|=|x-6|
=>3x-2=x-6 hoặc 3x-2=6-x
=>2x=4 hoặc 4x=8
=>x=2
c: |-21|-|5-x|=16
=>|x-5|=5
=>x-5=5 hoặc x-5=-5
=>x=10 hoặc x=0
\(-6x+2=44\Leftrightarrow x=\dfrac{42}{-6}=-7\)
\(3x-24+21-5x=27\Leftrightarrow-2x=30\Leftrightarrow x=-15\)
\(-11x+\left(5x+2\right)-8=36\)
\(\Leftrightarrow-11x+5x+2-8=36\)
\(\Leftrightarrow-6x=42\)
\(\Leftrightarrow x=-7\)
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\(3\left(x-8\right)+2x-5x=27\)
\(\Leftrightarrow3x-24+2x-5x=27\)
\(\Leftrightarrow-24=27\) (Vô lí)
Vậy không tìm được x thoả mãn đề bài.