Tớ làm câu a thui nha

a/ => (x³ + 27) - x(x² - 16) = 54

=> x³ + 27 - x³ + 16x = 54

=> 16x + 27 = 54

=> 16x = 27

=> x = 27/16

a) \(\left(x-5\right)\left(4-x\right)>0\)

\(\Rightarrow\left\{{}\begin{matrix}x-5>0\\4-x>0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x>5\\x>4\end{matrix}\right.\)\(\Rightarrow x>5\)

\(\left\{{}\begin{matrix}x-5< 0\\4-x< 0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x< 5\\x< 4\end{matrix}\right.\)\(\Rightarrow x< 4\)

Tập nghiệm: x > 5 ; x < 4

b) \(x^2-2x\ge0\)

\(\Leftrightarrow x\left(x-2\right)\ge0\)

\(\Rightarrow\left\{{}\begin{matrix}x\ge0\\x-2\ge0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x\ge0\\x\ge2\end{matrix}\right.\)\(\Rightarrow x\ge2\)

\(\left\{{}\begin{matrix}x\le0\\x-2\le0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x\le0\\x\le2\end{matrix}\right.\)\(\Rightarrow x\le0\)

Tập nghiệm: x >= 2 ; x<= 0

a) \(\left|x-\frac{3}{5}\right|=2x-\frac{2}{5}\)

\(\Leftrightarrow\orbr{\begin{cases}x-\frac{3}{5}=2x-\frac{2}{5}\\x-\frac{3}{5}=\frac{2}{5}-2x\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{5}\\3x=1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-\frac{1}{5}\\x=\frac{1}{3}\end{cases}}\)

Vậy \(x\in\left\{-\frac{1}{5};\frac{1}{3}\right\}\)

b) \(\left|x+0,37\right|=\left|2x-0,63\right|\)

\(\Leftrightarrow\orbr{\begin{cases}x+0,37=2x-0,63\\x+0,37=0,63-2x\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\3x=0,26\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{13}{150}\end{cases}}\)

Vậy \(x\in\left\{\frac{13}{150};1\right\}\)

help me plssssssssssss

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\(x^2+5y^2+z^2+2yz-12y+2x+10=0\)

\(\Leftrightarrow\left(x^2+2x+1\right)+\left(y^2+2yz+z^2\right)+\left(4y^2-12y+9\right)=0\)

\(\Leftrightarrow\left(x+1\right)^2+\left(y+z\right)^2+\left(2y-3\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+1=0\\y+z=0\\2y-3=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-1\\z=-\frac{3}{2}\\y=\frac{3}{2}\end{matrix}\right.\)

a,x^2+3x=0

=> x.(x+3)=0

=> +)x=0

+) x+3=0 => x=-3

b,x^3-4x=0

=> x.(x^2-2^2)=0

=> x.(x-2).(x+2)=0

=> +) x=0

+) x-2=0 => x=2

+) x+2=0 => x= -2

a) \(x^2+3x=0\)

\(x\left(x+3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x+3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-3\end{cases}}\)

                   vay \(\orbr{\begin{cases}x=0\\x=-3\end{cases}}\)

b) \(x^3-4x=0\)

\(x\left(x^2-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x^2-4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x^2=4\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

                           vay \(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

a) \(x^2+3x=0\)

   \(x\left(x+3\right)=0\)  

   \(\Rightarrow\orbr{\begin{cases}x=0\\x+3=0\end{cases}}\)

   \(\Rightarrow\orbr{\begin{cases}x=0\\x=-3\end{cases}}\)

Vậy \(x\in\left\{-3;0\right\}\)

b) \(x^3-4x\)

   \(x\left(x^2-2^2\right)=0\)

   \(x\left(x-2\right)\left(x+2\right)=0\)

=>     x = 0 

Hoặc x - 2 = 0   => x = 2

Hoặc x + 2 = 0  => x = -2

Vậy \(x\in\left\{-2;0;2\right\}\)

2,

|x+2| - |3x-1| = 0

⇒\(\left[{}\begin{matrix}x+2=0\\3x-1=0\end{matrix}\right.\)⇒\(\left[{}\begin{matrix}x=-2\\3x=1\end{matrix}\right.\)⇒\(\left[{}\begin{matrix}x=-2\\x=1:3\end{matrix}\right.\)

⇒\(\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy x ∈ \(\left\{\left(-2\right);\dfrac{1}{3}\right\}\)

1,

D= |x+1| + |x+3| + |x+5|

= |-x-1| + |x+3| + |x+5|

= |-x-1+x+3+x+5| = 1

Dấu bằng xảy ra khi -5 ≤ x ≤ -1

Vậy GTNN của D bằng 1 khi -5 ≤ x ≤ -1

Bài 1 mk lm bừa ko đúng đâu nha ☺

Tick mk bài 2 nhé

MẠI ZÔ MẠI ZÔ !!!

ta có \(\frac{x}{2}=\frac{y}{3};\frac{y}{4}=\frac{z}{5}\Rightarrow\frac{x}{8}=\frac{y}{12};\frac{y}{12}=\frac{z}{15}\Rightarrow\frac{x}{8}=\frac{y}{12}=\frac{z}{15}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\frac{x}{8}=\frac{y}{12}=\frac{z}{15}=\frac{x+y-z}{8+12-15}=\frac{10}{5}=2\)

Suy ra \(\frac{x}{8}=2\Rightarrow x=2.8=16\)

\(\frac{y}{12}=2\Rightarrow y=2.12=24\)

\(\frac{z}{15}=2\Rightarrow z=2.15=30\)

Vậy x=16;y=24;z=30