5

ko có?

\(x^2=-25\)

Vì \(x^2\ge0\forall x\)

Mà \(x^2=-25\) (vô lí)

Vậy: \(x\in\varnothing\)

\(x^2+5x=0\Leftrightarrow x\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

\(x^2-2x-xy+2y=\left(x^2-xy\right)-2\left(x-y\right)=x\left(x-y\right)-2\left(x-y\right)=\left(x-y\right)\left(x-2\right)\)

vì \(\left(4x^2-4x+1\right)^{2022}\ge0\left(\forall x\right)\),\(\left(y^2-\dfrac{4}{5}y+\dfrac{4}{25}\right)^{2022}\ge0\left(\forall y\right)\),\(\left|x+y+z\right|\ge0\)

mà \(\left(4x^2-4x+1\right)^{2022}+\left(y^2+\dfrac{4}{5}y+\dfrac{4}{25}\right)^{2022}+\left|x+y-z\right|=0\)

=>\(\left\{{}\begin{matrix}4x^2-4x+1=0\\y^2+\dfrac{4}{5}y+\dfrac{4}{25}=0\\x+y-z=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x-1=0\\y+\dfrac{2}{5}=0\\x+y-z=0\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{-2}{5}\\\dfrac{1}{2}-\dfrac{2}{5}-z=0\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{-2}{5}\\z=\dfrac{1}{10}\end{matrix}\right.\)

KL: vậy \(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{-2}{5}\\z=\dfrac{1}{10}\end{matrix}\right.\)

\(x^2-4x-4=25\)

\(\Leftrightarrow x^2-4x+4=33\)

\(\Leftrightarrow\left(x-2\right)^2=33\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=\sqrt{33}\\x-2=-\sqrt{33}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\sqrt{33}+2\\x=2-\sqrt{33}\end{cases}}\)

\(x^2-4x-4=25\)

\(\Rightarrow x^2-4x-4-25=0\)

\(\Rightarrow x^2-4x-29=0\)

\(\Rightarrow x^2-4x+4-33=0\)

\(\Rightarrow\left(x^2-4x+4\right)-33=0\)

\(\Rightarrow\left(x-2\right)^2=33\)

\(\Rightarrow\left(x-2\right)^2=\left(\sqrt{33}\right)^2=\left(-\sqrt{33}\right)^2\)

\(\Rightarrow\orbr{\begin{cases}x-2=\sqrt{33}\\x-2=-\sqrt{33}\end{cases}\Rightarrow\orbr{\begin{cases}x=2+\sqrt{33}\\x=2-\sqrt{33}\end{cases}}}\)

Vậy \(x\in\left\{2+\sqrt{33};2-\sqrt{33}\right\}\)

(x-2)² = 25 => x- 2 = 5
                  => x = 7

\(x^2-4x+4=25\)

\(\Leftrightarrow\left(x-2\right)^2=25\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=7\\x=-3\end{cases}}\)

a)

 ⇔ \(x^2-16=9\)

⇔ \(x^2=25\)

⇔ \(x=\pm5\)

b)

 ⇔ \(x^2-4x+4-25x^2+20x-4=0\)

⇔ \(16x-24x^2=0\)

⇔ \(8x\left(2-3x\right)=0\)

⇒ \(\left[{}\begin{matrix}x=0\\2-3x=0\end{matrix}\right.\)   ⇔   \(\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)

Vậy \(x=0\) hoặc \(x=\dfrac{2}{3}\)

c)  

⇔ \(3x^2-10x-20=0\)

⇔ \(x^2-2.x.\dfrac{5}{3}+\dfrac{25}{9}-\dfrac{205}{9}=0\)

⇔ \(\left(x-\dfrac{5}{3}\right)^2=\dfrac{205}{9}\)

⇒ \(\left[{}\begin{matrix}x-\dfrac{5}{3}=\sqrt{\dfrac{205}{9}}\\x-\dfrac{5}{3}=-\sqrt{\dfrac{205}{9}}\end{matrix}\right.\)  ⇔ \(\left[{}\begin{matrix}x=\dfrac{\sqrt{\text{205}}}{\text{3}}+\dfrac{5}{3}\\x=-\dfrac{\sqrt{\text{205}}}{\text{3}}+\dfrac{5}{3}\end{matrix}\right.\)  ⇔ \(\left[{}\begin{matrix}x=\dfrac{15+\text{9}\sqrt{\text{205}}}{\text{9}}\\\text{x}=-\dfrac{15+\text{9}\sqrt{\text{205}}}{\text{9}}\end{matrix}\right.\)

Vậy... 

d) 

⇔ \(\left(x^2+x\right)^2-49=\left(x^2+x\right)^2-7x\)

⇔ 7x = 49

⇔ x=7

Vậy...

1, \(\sqrt{4-4x+x^2}=3\)

\(\Leftrightarrow\sqrt{\left(2+x\right)^2}=3\)

\(\Leftrightarrow\left|2+x\right|=3\)

TH1: \(\left|2-x\right|=2-x\) với \(2-x\ge0\Leftrightarrow x\le2\)

Pt trở thành:

\(2-x=3\) (ĐK: \(x\le2\) )

\(\Leftrightarrow x=2-3\)

\(\Leftrightarrow x=-1\left(tm\right)\)

TH2: \(\left|2-x\right|=-\left(2-x\right)\) với \(2-x< 0\Leftrightarrow x>2\)

Pt trở thành:

\(-\left(2-x\right)=3\) (ĐK: \(x>2\))

\(\Leftrightarrow-2+x=3\)

\(\Leftrightarrow x=3+2\)

\(\Leftrightarrow x=5\left(tm\right)\)

Vậy \(S=\left\{-1;5\right\}\)

2, \(\sqrt{x^2-6x+9}=1\)

\(\Leftrightarrow\sqrt{x^2-2\cdot3\cdot x+3^2}=1\)

\(\Leftrightarrow\sqrt{\left(x-3\right)^2}=1\)

\(\Leftrightarrow\left|x-3\right|=1\)

TH1: \(\left|x-3\right|=x-3\) với \(x-3\ge0\Leftrightarrow x\ge3\)

Pt trở thành:

\(x-3=1\) (ĐK: \(x\ge3\))

\(\Leftrightarrow x=1+3\)

\(\Leftrightarrow x=4\left(tm\right)\)

TH2: \(\left|x-3\right|=-\left(x-3\right)\) với \(x-3< 0\Leftrightarrow x< 3\)

Pt trở thành:

\(-\left(x-3\right)=1\) (ĐK: \(x< 3\))

\(\Leftrightarrow-x+3=1\)

\(\Leftrightarrow-x=1-3\)

\(\Leftrightarrow-x=-2\)

\(\Leftrightarrow x=2\left(tm\right)\)

Vậy \(S=\left\{2;4\right\}\)

1) √(4 - 4x + x²) = 3

⇔ √(2 - x)² = 3

ĐKXĐ: Với mọi x ∈ R

⇔ |2 - x| = 3 (1)

*) |2 - x| = 2 - x ⇔ 2 - x ≥ 0 ⇔ x ≥ 2

(1) ⇔ 2 - x = 3

⇔ x = 2 - 3

⇔ x = -1 (nhận)

*) |2 - x| = x - 2 ⇔ 2 - x < 0 ⇔ x > 2

(1) ⇔ x - 2 = 3

⇔ x = 5 (nhận)

Vậy x = -1; x = 5

Q=3x+9y+15z+x+x4​+y+y9​+z+z25​

\ge 108+2.2+2.3+2.5=128≥108+2.2+2.3+2.5=128

Dấu "=" xảy ra khi x+3y+5z=36, x=\dfrac{4}x, y=\dfrac{9}y, z=\dfrac{25}z\Rightarrow x=2,y=3,z=5x+3y+5z=36,x=x4​,y=y9​,z=z25​⇒x=2,y=3,z=5

bạn tham khảo nhé