a) Ta có: 

a) Ta có: 

máy mình nó bị lỗi nên bn thông cảm nhé trả lời vừa đây mới là đugs

𝑎)2𝑥−1𝑥−3+4=−1𝑥−3

⇔2x-1x+1x=-3+3-4

⇔2x=-4

⇔x=-2

𝑏)3𝑥−22𝑥+5=6𝑥+14𝑥−3

⇔5+3=6x+14x-3x+22x

⇔8=39x

⇔x=\(\frac{8}{39}\)

𝑐)𝑥+3𝑥+1+𝑥−2𝑥=2

⇔x+3x+x-2x=2-1

⇔3x=1

⇔x=\(\frac{1}{3}\)

𝑑)x+1−2𝑥−3𝑥−1=2𝑥+3𝑥²−1

⇔3x²+2x+2x+3x-x-1-1+1=0

⇔3x²+6x-1=0

⇔3x²+3x+3x+3-4=0

⇔3x(x+1)+3(x+1)-4=0

⇔3(x+1)(x+1)-4=0

⇔3(x+1)²-4=0

⇔(x+1)²=\(\frac{4}{3}\)

⇔\(\left[{}\begin{matrix}x+1=\frac{4}{3}\\x+1=-\frac{4}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{4}{3}-1\\x=-\frac{4}{3}-1\end{matrix}\right.\)

Vậy ...

a, 2x - x - 3 + 4 = -x - 3

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

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

\(\Leftrightarrow\) 2x = -4

\(\Leftrightarrow\) x = -2

Vậy S = {-2}

b, 3x - 22x + 5 = 6x + 14x - 3

\(\Leftrightarrow\) -19x + 5 = 20x - 3

\(\Leftrightarrow\) -19x - 20x = -3 - 5

\(\Leftrightarrow\) -39x = -8

\(\Leftrightarrow\) x = \(\frac{8}{39}\)

Vậy S = {\(\frac{8}{39}\)}

c, x + 3x + 1 + x - 2x = 2

\(\Leftrightarrow\) 3x + 1 = 2

\(\Leftrightarrow\) 3x = 2 - 1

\(\Leftrightarrow\) 3x = 1

\(\Leftrightarrow\) x = \(\frac{1}{3}\)

Vậy S = {\(\frac{1}{3}\)}

Phần d mình ko hiểu, bạn viết rõ được ko!

Chúc bn học tốt!!

d, x + 1 - 2x - 3x - 1 = 2x + 3x² - 1

\(\Leftrightarrow\) x + 1 - 2x - 3x - 1 - 2x - 3x² + 1 = 0

\(\Leftrightarrow\) -3x² - 6x + 1 = 0

\(\Leftrightarrow\) -(3x² + 6x - 1) = 0

\(\Leftrightarrow\) 3x² + 6x - 1 = 0

\(\Leftrightarrow\) 3x² + 3x + 3x + 3 - 4 = 0

\(\Leftrightarrow\) 3x(x + 1) + 3(x + 1) - 4 = 0

\(\Leftrightarrow\) 3(x + 1)(x + 1) - 4 = 0

\(\Leftrightarrow\) 3(x + 1)² - 4 = 0

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

\(\Leftrightarrow\) x + 1 = \(\sqrt{\frac{4}{3}}\) hoặc x + 1 = \(-\sqrt{\frac{4}{3}}\)

\(\Leftrightarrow\) x = \(\sqrt{\frac{4}{3}}\) - 1 và x = \(-\sqrt{\frac{4}{3}}\) - 1

\(\Leftrightarrow\) x = \(\frac{2\sqrt{3}-3}{3}\) và x = \(\frac{-2\sqrt{3}-3}{3}\)

Vậy S = {\(\frac{2\sqrt{3}-3}{3}\); \(\frac{-2\sqrt{3}-3}{3}\)}

Chúc bn học tốt!!

Nguyễn Thị Anh Thư cái này bạn gửi một lần r mà!

Bài 1:

b: Tọa độ giao điểm là:

\(\left\{{}\begin{matrix}-\dfrac{1}{2}x+3=x-2\\y=x-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{2}x=5\\y=x-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{10}{3}\\y=\dfrac{10}{3}-2=\dfrac{4}{3}\end{matrix}\right.\)

a: Ta có: \(A=-x^2+2x+5\)

\(=-\left(x^2-2x-5\right)\)

\(=-\left(x^2-2x+1-6\right)\)

\(=-\left(x-1\right)^2+6\le6\forall x\)

Dấu '=' xảy ra khi x=1

b: Ta có: \(B=-x^2-8x+10\)

\(=-\left(x^2+8x-10\right)\)

\(=-\left(x^2+8x+16-26\right)\)

\(=-\left(x+4\right)^2+26\le26\forall x\)

Dấu '=' xảy ra khi x=-4

c: Ta có: \(C=-3x^2+12x+8\)

\(=-3\left(x^2-4x-\dfrac{8}{3}\right)\)

\(=-3\left(x^2-4x+4-\dfrac{20}{3}\right)\)

\(=-3\left(x-2\right)^2+20\le20\forall x\)

Dấu '=' xảy ra khi x=2

d: Ta có: \(D=-5x^2+9x-3\)

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

\(=-5\left(x^2-2\cdot x\cdot\dfrac{9}{10}+\dfrac{81}{100}-\dfrac{21}{100}\right)\)

\(=-5\left(x-\dfrac{9}{10}\right)^2+\dfrac{21}{20}\le\dfrac{21}{20}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{9}{10}\)

e: Ta có: \(E=\left(4-x\right)\left(x+6\right)\)

\(=4x+24-x^2-6x\)

\(=-x^2-2x+24\)

\(=-\left(x^2+2x-24\right)\)

\(=-\left(x^2+2x+1-25\right)\)

\(=-\left(x+1\right)^2+25\le25\forall x\)

Dấu '=' xảy ra khi x=-1

f: Ta có: \(F=\left(2x+5\right)\left(4-3x\right)\)

\(=8x-6x^2+20-15x\)

\(=-6x^2-7x+20\)

\(=-6\left(x^2+\dfrac{7}{6}x-\dfrac{10}{3}\right)\)

\(=-6\left(x^2+2\cdot x\cdot\dfrac{7}{12}+\dfrac{49}{144}-\dfrac{529}{144}\right)\)

\(=-6\left(x+\dfrac{7}{12}\right)^2+\dfrac{529}{24}\le\dfrac{529}{24}\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{7}{12}\)

\(a,=\left(4x^2-1\right)\left(2x-5\right)=8x^3-20x^2-2x+5\\ b,=\left[x^2+\left(x-3\right)\right]\left[x^2-\left(x-3\right)\right]=x^4-\left(x-3\right)^2\\ =x^4-x^2+6x-9\)