3x-4=26
giúp mk với
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a,ĐKXĐ:\(x\ge2\)
\(4\sqrt{x-2}+\sqrt{9x-18}-\sqrt{\dfrac{x-2}{4}}=26\\ \Leftrightarrow4\sqrt{x-2}+3\sqrt{x-2}-\dfrac{\sqrt{x-2}}{2}=26\\ \Leftrightarrow8\sqrt{x-2}+6\sqrt{x-2}-\sqrt{x-2}=52\\ \Leftrightarrow13\sqrt{x-2}=52\\ \Leftrightarrow\sqrt{x-2}=4\\ \Leftrightarrow x-2=16\\ \Leftrightarrow x=18\left(tm\right)\)
b,ĐKXĐ:\(x\in R\)
\(3x+\sqrt{4x^2-8x+4}=1\\ \Leftrightarrow2\sqrt{x^2-2x+1}=1-3x\\ \Leftrightarrow\left|x-1\right|=\dfrac{1-3x}{2}\\ \Leftrightarrow\left[{}\begin{matrix}x-1=\dfrac{1-3x}{2}\\x-1=\dfrac{3x-1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x-2=1-3x\\2x-2=3x-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)
c, ĐKXĐ:\(x\ge0\)
\(\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)=7\\ \Leftrightarrow\sqrt{x}\left(2\sqrt{x}+1\right)-2\left(2\sqrt{x}+1\right)=7\\ \Leftrightarrow2x+\sqrt{x}-4\sqrt{x}-2=7\\ \Leftrightarrow2x-3\sqrt{x}-9=0\\ \Leftrightarrow\left(2x+3\sqrt{x}\right)-\left(6\sqrt{x}+9\right)=0\\ \Leftrightarrow\sqrt{x}\left(2\sqrt{x}+3\right)-3\left(2\sqrt{x}+3\right)=0\\ \Leftrightarrow\left(\sqrt{x}-3\right)\left(2\sqrt{x}+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=3\\2\sqrt{x}=-3\left(vô.lí\right)\end{matrix}\right.\\ \Leftrightarrow x=9\left(tm\right)\)
\(\dfrac{3}{5}\) x 0,25\(x\) = - \(\dfrac{1}{2}\)
0,25\(x\) = - \(\dfrac{1}{2}\) : \(\dfrac{3}{5}\)
0,25\(x\) = - \(\dfrac{1}{2}\) x \(\dfrac{5}{3}\)
0,25\(x\) = - \(\dfrac{5}{6}\)
\(x\) = - \(\dfrac{5}{6}\) : 0,25
\(x\) = - \(\dfrac{5}{6}\) x 4
\(x\) = - \(\dfrac{10}{3}\)
Vậy \(x\) = - \(\dfrac{10}{3}\)
2626 : (\(\dfrac{1}{2}\)\(x\) + \(\dfrac{5}{2}\)\(x\)) = 26
\(\dfrac{1}{2}\)\(x\) + \(\dfrac{5}{2}\)\(x\) = 2626 : 26
\(\dfrac{1}{2}\)\(x\) + \(\dfrac{5}{2}\)\(x\) = 101
\(x\) x ( \(\dfrac{1}{2}\) + \(\dfrac{5}{2}\)) = 101
\(x\) x 3 = 101
\(x\) = 101 : 3
\(x\) = \(\dfrac{101}{3}\)
Vậy \(x\) = \(\dfrac{101}{3}\)
\(5x^2+3x-26=0\)
\(\Leftrightarrow5x^2+13x-10x-26=0\)
\(\Leftrightarrow x\left(5x+13\right)-2\left(5x+13\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(5x+13\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-2=0\\5x+13=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=-\frac{13}{5}\end{array}\right.\)
\(\Leftrightarrow9x+3=-20x+16\)
=>29x=13
hay x=13/29
\(\Leftrightarrow\dfrac{3x-1}{4}-\dfrac{3x-6}{8}-\dfrac{5-3x}{2}>1\)
\(\Leftrightarrow\dfrac{\left(3x-1\right).2-\left(3x-6\right)-\left(5-3x\right).4}{8}>1\)
\(\Leftrightarrow\dfrac{6x-2-3x+6-20+12x}{8}>1\)
<=> 15x - 16 > 8
<=> 15x > 24
<=> x > 8/5
Ta có: \(\dfrac{3x-1}{4}-\dfrac{3\left(x-2\right)}{8}-1>\dfrac{5-3x}{2}\)
\(\Leftrightarrow2\left(3x-1\right)-3\left(x-2\right)-8>4\left(5-3x\right)\)
\(\Leftrightarrow6x-2-3x+6-8>20-12x\)
\(\Leftrightarrow3x-4-20+12x>0\)
\(\Leftrightarrow15x>24\)
hay \(x>\dfrac{8}{5}\)
\(\dfrac{3x-1}{4}-\dfrac{3\left(x-2\right)}{8}-1>\dfrac{5-3x}{2}\)
MTC : 8
\(\Rightarrow\dfrac{2\left(3x-1\right)}{8}-\dfrac{3\left(x-2\right)}{8}-\dfrac{8}{8}>\dfrac{4\left(5-3x\right)}{8}\)
Suy ra : 2(3x - 1) - 3(x - 2) - 8 > 4(5 - 3x)
\(\Leftrightarrow\) 6x - 2 - 3x + 6 - 8 > 20 - 12x
\(\Leftrightarrow\) 6x - 3x + 12x > 20 + 2 - 6 + 8
\(\Leftrightarrow\) 15x > 24
\(\Leftrightarrow\) x > \(\dfrac{24}{15}=\dfrac{8}{5}\)
Vay x >\(\dfrac{8}{5}\)
Chuc ban hoc tot
a) (x + 4) + (x + 6) + (x + 8) + ... + (x + 26) = 210
Các phần tử của phép tính trên là: (26 - 4) : 2 + 1 = 12(phần tử) ⇒ 12x
Tổng của phép tính trên là: (26 + 4) x 12 : 2 = 180
Tính:
(x + 4) + (x + 6) + (x + 8) + ... + (x + 26) = 210
12x + (4 + 6 + 8 + ... + 26) = 210
12x + 180 = 210
12x = 210 - 180
12x = 30
x = 30 : 12
x = 2,5
a) (x + 4) + (x + 6) + (x + 8) + ... + (x + 26) = 210
Các phần tử của phép tính trên là: (26 - 4) : 2 + 1 = 12(phần tử) ⇒ 12x
Tổng của phép tính trên là: (26 + 4) x 12 : 2 = 180
Tính:
(x + 4) + (x + 6) + (x + 8) + ... + (x + 26) = 210
12x + (4 + 6 + 8 + ... + 26) = 210
12x + 180 = 210
12x = 210 - 180
12x = 30
x = 30 : 12
x = 2,5
b) x + 280 : 25 - 7,2 = 15
x + 11,2 - 7,2 = 15
x + 11,2 = 15 + 7,2
x + 11,2 = 22,2
x = 22,2 - 11,2
x = 11
ĐKXĐ: \(-1\le x\le\dfrac{5}{2}\)
\(\Leftrightarrow\sqrt{3x+3}-3+1-\sqrt{5-2x}=x^3-3x^2-10x+24\)
\(\Leftrightarrow\dfrac{3\left(x-2\right)}{\sqrt{3x+3}+3}+\dfrac{2\left(x-2\right)}{1+\sqrt{5-2x}}=\left(x-2\right)\left(x-4\right)\left(x+3\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\\dfrac{3}{\sqrt{3x+3}+3}+\dfrac{2}{1+\sqrt{5-2x}}=\left(x-4\right)\left(x+3\right)\left(1\right)\end{matrix}\right.\)
Xét (1), ta có:
\(\dfrac{3}{\sqrt{3x+3}+3}+\dfrac{2}{1+\sqrt{5-2x}}>0\)
\(-1\le x\le\dfrac{5}{2}\Rightarrow\left\{{}\begin{matrix}x+3>0\\x-4< 0\end{matrix}\right.\) \(\Rightarrow\left(x+3\right)\left(x-4\right)< 0\)
\(\Rightarrow\left(1\right)\) vô nghiệm hay pt có nghiệm duy nhất \(x=2\)
`3x-4=26`
`=>3x=26+4`
`=>3x=30`
`=>x=30/3`
`=>x=10`
Vậy: ...
\(3x-4=26\Leftrightarrow3x=30\Leftrightarrow x=10\)