\(x^2-x+\dfrac{1}{4}=0\)

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=0\)

\(\Leftrightarrow x-\dfrac{1}{2}=0\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

Vậy \(x=\dfrac{1}{2}\)

Wish you study well !!

x+1/x+2+4/4-x^2=1-5x^2+10x+20/x^3-8

nhầm rồi

1cong1=

Vì x, y > 0

Đặt \(\frac{x}{5}=\frac{y}{4}=k\Rightarrow\hept{\begin{cases}x=5k\\y=4k\end{cases}}\)( k > 0 ) 

x² - y² = 4

<=> ( 5k )² - ( 4k )² = 4

<=> 25k² - 16k² = 4

<=> 9k² = 4

<=> k² = 4/9

<=> k = 2/3 ( vì k > 0 )

=> \(\hept{\begin{cases}x=5\cdot\frac{2}{3}=\frac{10}{3}\\y=4\cdot\frac{2}{3}=\frac{8}{3}\end{cases}}\)

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Ta có x/5=y/5=k
=>x=5k;y=4k
Theo tính chất dãy tỉ số = nhau ta có
x/5=y/4=x^2-y^2/5^2-4^2=4/9
x/5=4/9=>x=4/9.5=20/9
y/4=4/9=>y=4/9.4=16/9
=>x=20/9;y=16/9

a) -9-25=(7-x)-(25+7)

<=> −9−25=7−x−25−7

<=> -9=7-x-7

<=> -9=-x

<=> x=9

b) (x-5)(x+8)=0

<=> x-5=0 hoặc x+8=0

<=> x=5 hoặc x=-8

d) Đề sai, x đâu???

...

\(c,0< \left|x-7\right|< 3\)

\(0< x-7< 3\)

\(7< x< 10\)

\(\Rightarrow x\in\left\{8;9\right\}\)

\(x^3-x=0\Rightarrow x\left(x^2-1\right)=0\)

TH1: \(x=0\)

TH2: \(x^2-1=0\Rightarrow x^2=1\Rightarrow x=\sqrt{1}\)hoặc \(x=-\sqrt{1}\)

\(x.2-5.x=0\)

\(\Rightarrow-3x=0\)

\(\Rightarrow x=0\)

Vậy \(x=0\)

\(2x-5x=0\)

\(=>2x=5x\)

\(=>x=0\)

Trả lời đúng , mình sẽ cho 5 vote

\(\Rightarrow x-\left(-x+x+3\right)-\left(x+3-x+2\right)=0\\ \Rightarrow x-3-5=0\\ \Rightarrow x=8\)

Sai rồi bạn ơi, bằng -2 mới đúng,bn thử lại coi