a, 4x² - 49 = 0

⇔⇔ (2x)² - 7² = 0

⇔⇔ (2x - 7)(2x + 7) = 0

⇔{2x−7=02x+7=0⇔⎧⎪ ⎪⎨⎪ ⎪⎩x=72x=−72⇔{2x−7=02x+7=0⇔{x=72x=−72

b, x² + 36 = 12x

⇔⇔ x² + 36 - 12x = 0

⇔⇔ x² - 2.x.6 + 6² = 0

⇔⇔ (x - 6)² = 0

⇔⇔ x = 6

e, (x - 2)² - 16 = 0

⇔⇔ (x - 2)² - 4² = 0

⇔⇔ (x - 2 - 4)(x - 2 + 4) = 0

⇔⇔ (x - 6)(x + 2) = 0

⇔{x−6=0x+2=0⇔{x=6x=−2⇔{x−6=0x+2=0⇔{x=6x=−2

f, x² - 5x -14 = 0

⇔⇔ x² + 2x - 7x -14 = 0

⇔⇔ x(x + 2) - 7(x + 2) = 0

⇔⇔ (x + 2)(x - 7) = 0

⇔{x+2=0x−7=0⇔{x=−2x=7

a,\(4x^2-49=0\)

\(\Leftrightarrow\left(2x\right)^2-7^2=0\)

\(\Leftrightarrow\left(2x-7\right)\left(2x+7\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x-7=0\\2x+7=0\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=7\\2x=-7\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{7}{2}\end{cases}}}\)

b.\(x^2+36=12x\)

\(\Leftrightarrow x^2-12x+36=0\)

\(\Leftrightarrow\left(x-6\right)^2=0\)

\(\Leftrightarrow x-6=0\Leftrightarrow x=6\)

c.\(\frac{1}{16x^2}-x+4=0\)

\(\Leftrightarrow\left(\frac{1}{4x}\right)^2-2.\frac{1}{4x}.2+2^2=0\)

\(\Leftrightarrow\left(\frac{1}{4x}-2\right)^2=0\)

........

Tham khảo thanh này để soạn đề chính xác hơn nha :vvv

a) Ta có: \(M=\left(\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\right)\cdot\dfrac{x+3\sqrt{x}}{7-\sqrt{x}}\)

\(=\left(\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right)\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{7-\sqrt{x}}\)

\(=\dfrac{x-9-\left(x-2\sqrt{x}+\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{7-\sqrt{x}}\)

\(=\dfrac{x-9-x+\sqrt{x}+2}{\left(\sqrt{x}-2\right)}\cdot\dfrac{1}{-\left(\sqrt{x}-7\right)}\)

\(=\dfrac{\sqrt{x}-7}{\sqrt{x}-2}\cdot\dfrac{-1}{\sqrt{x}-7}\)

\(=\dfrac{-1}{\sqrt{x}-2}\)(1)

b) Ta có: \(x^2-4x=0\)

\(\Leftrightarrow x\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=4\left(loại\right)\end{matrix}\right.\)

Thay x=0 vào biểu thức (1), ta được:

\(M=\dfrac{-1}{\sqrt{0}-2}=\dfrac{-1}{-2}=\dfrac{1}{2}\)

Vậy: Khi \(x^2-4x=0\) thì \(M=\dfrac{1}{2}\)

a: \(x\in\left\{0;25\right\}\)

c: \(x\in\left\{0;5\right\}\)

x(x+2)=0

suy ra x=0 hoặc x+2=0

5-2x=-7

2x=-7+5

2x=-(7-5)

2x=-2

x=-2:2

x=-1

Vậy x=-1

NHỚ TÍCH MK NHA

Tự học giúp bạn có được một gia tài
Jim Rohn – Triết lý cuộc đời

là sao?

a) \(\frac{3}{4}-\left(\frac{1}{2}:x+\frac{1}{2}\right)=\frac{3}{5}\)

\(\Leftrightarrow\frac{1}{2}:x+\frac{1}{2}=\frac{3}{4}-\frac{3}{5}\)

\(\Leftrightarrow\frac{1}{2}:x+\frac{1}{2}=\frac{15}{20}-\frac{12}{20}\)

\(\Leftrightarrow\frac{1}{2}:x+\frac{1}{2}=\frac{13}{20}\)

\(\Leftrightarrow\frac{1}{2}:x=\frac{13}{20}-\frac{1}{2}\)

\(\Leftrightarrow\frac{1}{2}:x=\frac{13}{20}-\frac{10}{20}\)

\(\Leftrightarrow\frac{1}{2}:x=\frac{3}{20}\)

\(\Leftrightarrow x=\frac{1}{2}:\frac{3}{20}\)

\(\Leftrightarrow x=\frac{1}{2}.\frac{20}{3}=\frac{10}{3}\)

Vậy: \(x=\frac{10}{3}\)

b) \(3x.\left(\frac{1}{2}.x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x=0\\\frac{1}{2}x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\\frac{1}{2}x=1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\x=1:\frac{1}{2}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\x=2\end{cases}}}\)

Vậy: \(x\in\left\{0;2\right\}\)

c) \(\left(4-x\right)\left(2x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}4-x=0\\2x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\2x=3\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=4\\x=\frac{3}{2}\end{cases}}}\)

Vậy: \(x\in\left\{4;\frac{3}{2}\right\}\)

d) \(\frac{4}{-3}=\frac{-12}{x}\)

\(\Leftrightarrow4x=\left(-12\right).\left(-3\right)\)

\(\Leftrightarrow4x=36\)

\(\Leftrightarrow x=9\)

Vậy: \(x=9\)

e) \(\frac{4x}{-3}=\frac{12}{-x}\)

\(\Leftrightarrow4x.\left(-x\right)=12.\left(-3\right)\)

\(\Leftrightarrow-4x^2=-36\)

\(\Leftrightarrow x^2=9\)

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

Vậy: \(x\in\left\{3;-3\right\}\)

a,Ta có: 3/4-(1/2:x+1/2)=3/5

                  -(1/2:x+1/2)=3/5-3/4

                  -(1/2:x+1/2)=-3/20

                      1/2:x+1/2=3/20

                             1/2:x=3/20-1/2

                             1/2:x=-7/20

                                   x=1/2:-7/20

                                   x=-10/7

b,Ta có: 3x.(1/2x-1)=0

 Với 3x=0 =>x=0

vói1/2x-1=0

a)\(\frac{3}{4}-\left(\frac{1}{2}:x+\frac{1}{2}\right)=\frac{3}{5}\)

\(\Leftrightarrow\frac{1}{2}:x+\frac{1}{2}=\frac{3}{20}\)

\(\Leftrightarrow\frac{1}{2}:x=\frac{-7}{20}\Leftrightarrow x=-\frac{10}{7}\)

b) \(3x\left(\frac{1}{2}x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x=0\\\frac{1}{2}x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\\frac{1}{2}x=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

Vậy x = 0 hoặc x = 2

c) \(\left(4-x\right)\left(2x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}4-x=0\\2x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=4\\2x=-3\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=4\\x=-\frac{3}{2}\end{cases}}\)

Vậy x = 4 hoặc x = -3/2

d) \(\frac{4}{-3}=-\frac{12}{x}\Leftrightarrow4x=36\Leftrightarrow x=9\)

e) \(\frac{4x}{-3}=\frac{12}{-x}\)

\(\Leftrightarrow4x\left(-x\right)=-36\)

\(\Leftrightarrow x\left(-x\right)=-9\)

\(\Leftrightarrow-x^2=-9\)

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

Chia nhỏ ra

a: =>1/2x=7/2-2/3=21/6-4/6=17/6

=>x=17/3

b: =>2/3:x=-7-1/3=-22/3

=>x=2/3:(-22/3)=-1/11

c: =>1/3x+2/5x-2/5=0

=>11/15x=2/5

hay x=6/11

d: =>2x-3=0 hoặc 6-2x=0

=>x=3/2 hoặc x=3

a) Ta có: \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=15\)

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6\left(x^2+2x+1\right)=15\)

\(\Leftrightarrow-6x^2+12x+19+6x^2+12x+6=15\)

\(\Leftrightarrow24x+25=15\)

\(\Leftrightarrow24x=-10\)

hay \(x=-\dfrac{5}{12}\)

b) Ta có: \(2x^3-50x=0\)

\(\Leftrightarrow2x\left(x-5\right)\left(x+5\right)=0\)

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

c) Ta có: \(5x^2-4\left(x^2-2x+1\right)-5=0\)

\(\Leftrightarrow5x^2-4x^2+8x-4-5=0\)

\(\Leftrightarrow x^2+8x-9=0\)

\(\Leftrightarrow\left(x+9\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=1\end{matrix}\right.\)

d) Ta có: \(x^3-x=0\)

\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)

e) Ta có: \(27x^3-27x^2+9x-1=1\)

\(\Leftrightarrow\left(3x\right)^3-3\cdot\left(3x\right)^2\cdot1+3\cdot3x\cdot1^2-1^3=1\)

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

\(\Leftrightarrow3x-1=1\)

\(\Leftrightarrow3x=2\)

hay \(x=\dfrac{2}{3}\)

d,

\(|x-\frac{1}{3}|=\frac{5}{6}\Rightarrow \left[\begin{matrix} x-\frac{1}{3}=\frac{5}{6}\\ x-\frac{1}{3}=-\frac{5}{6}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{7}{6}\\ x=\frac{-1}{2}\end{matrix}\right.\)

e,

\(\frac{3}{4}-2|2x-\frac{2}{3}|=2\)

\(\Leftrightarrow 2|2x-\frac{2}{3}|=\frac{3}{4}-2=\frac{-5}{4}\)

\(\Leftrightarrow |2x-\frac{2}{3}|=-\frac{5}{8}<0\) (vô lý vì trị tuyệt đối của 1 số luôn không âm)

Vậy không tồn tại $x$ thỏa mãn đề bài.

f, 

\(\frac{2x-1}{2}=\frac{5+3x}{3}\Leftrightarrow 3(2x-1)=2(5+3x)\)

\(\Leftrightarrow 6x-3=10+6x\)

\(\Leftrightarrow 13=0\) (vô lý)

Vậy không tồn tại $x$ thỏa mãn đề bài.

a,

$0-|x+1|=5$

$|x+1|=0-5=-5<0$ (vô lý do trị tuyệt đối của một số luôn không âm)

Do đó không tồn tại $x$ thỏa mãn điều kiện đề.

b,

\(2-|\frac{3}{4}-x|=\frac{7}{12}\)

\(|\frac{3}{4}-x|=2-\frac{7}{12}=\frac{17}{12}\)

\(\Rightarrow \left[\begin{matrix} \frac{3}{4}-x=\frac{17}{12}\\ \frac{3}{4}-x=\frac{-17}{12}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{-2}{3}\\ x=\frac{13}{6}\end{matrix}\right.\)

c, 

\(2|\frac{1}{2}x-\frac{1}{3}|-\frac{3}{2}=\frac{1}{4}\)

\(2|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{4}\)

\(|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{8}\)

\(\Rightarrow \left[\begin{matrix} \frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\ \frac{1}{2}x-\frac{1}{3}=-\frac{7}{8}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{29}{12}\\ x=\frac{-13}{12}\end{matrix}\right.\)