Cậu tách ra `2->3` câu thôi nhe

a: =>17x-5x-15-2x-5=0

=>10x-20=0

=>x=2

b: =>\(\dfrac{3x-6-5x-10}{\left(x+2\right)\left(x-2\right)}=\dfrac{11x+23}{\left(x+2\right)\left(x-2\right)}\)

=>11x+23=-2x-16

=>13x=-39

=>x=-3(nhận)

c: =>5x+7>=3x-3

=>2x>=-10

=>x>=-5

d: =>5(3x-1)=-2(x+1)

=>15x-5=-2x-2

=>17x=3

=>x=3/17

e: =>4x^2-1-4x^2-3x-2=0

=>-3x-3=0

=>x=-1

g: =>7x-5-8x+2-7<0

=>-x-10<0

=>x+10>0

=>x>-10

a: =>17x-5x-15-2x-5=0

=>10x-20=0

=>x=2

b: =>\(\dfrac{3x-6-5x-10}{\left(x+2\right)\left(x-2\right)}=\dfrac{11x+23}{\left(x+2\right)\left(x-2\right)}\)

=>11x+23=-2x-16

=>13x=-39

=>x=-3(nhận)

c: =>5x+7>=3x-3

=>2x>=-10

=>x>=-5

d: =>5(3x-1)=-2(x+1)

=>15x-5=-2x-2

=>17x=3

=>x=3/17

e: =>4x^2-1-4x^2-3x-2=0

=>-3x-3=0

=>x=-1

g: =>7x-5-8x+2-7<0

=>-x-10<0

=>x+10>0

=>x>-10

ta có : x^5+2x^4+3x^3+3x^2+2x+1=0

\(\Leftrightarrow\)x^5+x^4+x^4+x^3+2x^3+2x^2+x^2+x+x+1=0

\(\Leftrightarrow\)(x^5+x^4)+(x^4+x^3)+(2x^3+2x^2)+(x^2+x)+(x+1)=0

\(\Leftrightarrow\)x^4(x+1)+x^3(x+1)+2x^2(x+1)+x(x+1)+(x+1)=0

\(\Leftrightarrow\)(x+1)(x^4+x^3+2x^2+x+1)=0

\(\Leftrightarrow\)(x+1)(x^4+x^3+x^2+x^2+x+1)=0

\(\Leftrightarrow\)(x+1)[x^2(x^2+x+1)+(x^2+x+1)]=0

\(\Leftrightarrow\)(x+1)(x^2+x+1)(x^2+1)=0

VÌ x^2+x+1=(x+\(\dfrac{1}{2}\))^2+\(\dfrac{3}{4}\)\(\ne0\) và x^2+1\(\ne0\)

\(\Rightarrow\)x+1=0

\(\Rightarrow\)x=-1

CÒN CÂU B TỰ LÀM (02042006)

b: x^4+3x^3-2x^2+x-3=0

=>x^4-x^3+4x^3-4x^2+2x^2-2x+3x-3=0

=>(x-1)(x^3+4x^2+2x+3)=0

=>x-1=0

=>x=1

a.

ĐKXĐ: \(x\ge-\dfrac{5}{3}\)

\(9x^2-3x-\left(3x+5\right)-\sqrt{3x+5}=0\)

Đặt \(\sqrt{3x+5}=t\ge0\)

\(\Rightarrow9x^2-3x-t^2-t=0\)

\(\Delta=9+36\left(t^2+t\right)=\left(6t+3\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3+6t+3}{18}=\dfrac{t+1}{3}\\x=\dfrac{3-6t-3}{18}=-\dfrac{t}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}t=3x-1\\t=-3x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{3x+5}=3x-1\left(x\ge\dfrac{1}{3}\right)\\\sqrt{3x+5}=-3x\left(x\le0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+5=9x^2-6x+1\left(x\ge\dfrac{1}{3}\right)\\3x+5=9x^2\left(x\le0\right)\end{matrix}\right.\)

\(\Leftrightarrow...\)

c.

ĐKXĐ: \(x\ge-5\)

\(x^2-3x+2-x-5-\sqrt{x+5}=0\)

Đặt \(\sqrt{x+5}=t\ge0\)

\(\Rightarrow-t^2-t+x^2-3x+2=0\)

\(\Delta=1+4\left(x^2-3x+2\right)=\left(2x-3\right)^2\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+5}=1-x\left(x\le1\right)\\\sqrt{x+5}=x-2\left(x\ge2\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=x^2-2x+1\left(x\le1\right)\\x+5=x^2-4x+4\left(x\ge2\right)\end{matrix}\right.\)

\(\Leftrightarrow...\)

b.

ĐKXĐ: \(x\ge-\dfrac{8}{3}\)

\(\left(3x+2\right)^2-6-\sqrt{3x+8}=0\)

Đặt \(\sqrt{3x+8}=t\ge0\Rightarrow3x+2=t^2-6\)

\(\left(t^2-6\right)^2-6-t=0\)

\(\Leftrightarrow t^4-12t^2-t+30=0\)

\(\Leftrightarrow\left(t^2+t-5\right)\left(t^2-t-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=3\\t=\dfrac{\sqrt{21}-1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{3x+8}=3\\\sqrt{3x+8}=\dfrac{\sqrt{21}-1}{2}\end{matrix}\right.\)

\(\Leftrightarrow...\)

a) \(x^5+2x^4+3x^3+3x^2+2x+1=0\)

\(\Leftrightarrow x^5+x^4+x^4+x^3+2x^3+2x^2+x^2+x+x+1=0\)

\(\Leftrightarrow x^4\left(x+1\right)+x^3\left(x+1\right)+2x^2\left(x+1\right)+x\left(x+1\right)+\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^4+x^3+2x^2+x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^4+x^3+x^2+x^2+x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left[x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)\right]=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2+x+1\right)\left(x^2+1\right)=0\)

Dễ thấy \(x^2+x+1>0\forall x;x^2+1>0\forall x\)

\(\Rightarrow x+1=0\)

\(\Leftrightarrow x=-1\)

Vậy....

b) \(x^4+3x^3-2x^2+x-3=0\)

\(\Leftrightarrow x^4-x^3+4x^3-4x^2+2x^2-2x+3x-3=0\)

\(\Leftrightarrow x^3\left(x-1\right)+4x^2\left(x-1\right)+2x\left(x-1\right)+3\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^3+4x^2+2x+3\right)=0\)

...

\(\Leftrightarrow x=1\)

p/s: có bác nào giải đc pt \(x^3+4x^2+2x+3=0\)thì giúp nhé :))

1.a)|−7x|=3x+16

Vì |-7x| ≥ 0  nên 3x+16 ≥ 0 ⇔ x ≥ \(\dfrac{-16}{3}\)    (*)

Với đk (*), ta có: |-7x|=3x+16

\(\left[\begin{array}{} -7x=3x+16\\ -7x=-3x-16 \end{array} \right.\) ⇔  \(\left[\begin{array}{} -7x-3x=16\\ -7x+3x=-16 \end{array} \right.\)

⇔ \(\left[\begin{array}{} x=-1,6 (t/m)\\ x= 4 (t/m) \end{array} \right.\)

b) \(\dfrac{x-1}{x+2}\) - \(\dfrac{x}{x-2}\) = \(\dfrac{5x-8}{x^2-4}\)

⇔ \(\dfrac{(x-1)(x-2)}{x^2-4}\) - \(\dfrac{x(x+2)}{x^2-4}\) = \(\dfrac{5x-8}{x^2-4}\)

⇒ x² - 2x - x + 2 - x² - 2x = 5x - 8  

⇔ -5x - 5x = -8 - 2

⇔ -10x = -10

⇔ x=1

2.7x+5 < 3x−11

⇔ 7x - 3x < -11 - 5

⇔ 4x < -16

⇔ x < -4

bạn tự biểu diễn trên trục số nha !

a.

\(3\sqrt[3]{3\left(x+1\right)+2}=\left(x+1\right)^3-2\)

Đặt \(\sqrt[3]{3\left(x+1\right)+2}=y\) ta được:

\(\left\{{}\begin{matrix}3y=\left(x+1\right)^3-2\\3\left(x+1\right)+2=y^3\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}3y+2=\left(x+1\right)^3\\3\left(x+1\right)+2=y^3\end{matrix}\right.\)

\(\Rightarrow\left(x+1\right)^3-y^3=3y-3\left(x+1\right)\)

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

\(\Leftrightarrow x+1=y\)

\(\Leftrightarrow\left(x+1\right)^3=y^3\)

\(\Leftrightarrow\left(x+1\right)^3=3\left(x+1\right)+2\)

\(\Leftrightarrow x^3+3x^2-4=0\)

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

b.

\(\Leftrightarrow8x^3-\left(6x+1\right)+2x-\sqrt[3]{6x+1}=0\)

Đặt \(\left\{{}\begin{matrix}2x=a\\\sqrt[3]{6x+1}=b\end{matrix}\right.\) ta được:

\(a^3-b^3+a-b=0\)

\(\Leftrightarrow\left(a-b\right)\left(a^2+ab+b^2+1\right)=0\)

\(\Leftrightarrow a=b\)

\(\Leftrightarrow2x=\sqrt[3]{6x+1}\)

\(\Leftrightarrow8x^3-6x-1=0\)

Đặt \(f\left(x\right)=8x^3-6x-1\)

\(f\left(x\right)\) là hàm đa thức nên liên tục trên R, đồng thời \(f\left(x\right)\) bậc 3 nên có tối đa 3 nghiệm

\(f\left(-1\right)=-3< 0\) ; \(f\left(-\dfrac{1}{2}\right)=1>0\) \(\Rightarrow f\left(-1\right).f\left(-\dfrac{1}{2}\right)< 0\)

\(\Rightarrow f\left(x\right)\) có 1 nghiệm thuộc \(\left(-1;-\dfrac{1}{2}\right)\) (1)

\(f\left(0\right)=-1\Rightarrow f\left(0\right).f\left(-\dfrac{1}{2}\right)< 0\Rightarrow f\left(x\right)\) có 1 nghiệm thuộc \(\left(-\dfrac{1}{2};0\right)\) (2)

\(f\left(1\right)=1\Rightarrow f\left(0\right).f\left(1\right)< 0\Rightarrow f\left(x\right)\) có 1 nghiệm thuộc \(\left(0;1\right)\) (3)

Từ (1);(2);(3) \(\Rightarrow\) cả 3 nghiệm của \(f\left(x\right)\) đều thuộc \(\left(-1;1\right)\)

Do đó, ta chỉ cần tìm nghiệm của \(f\left(x\right)\) với \(x\in\left(-1;1\right)\)

Do \(x\in\left(-1;1\right)\), đặt \(x=cosu\)

\(\Rightarrow8cos^3u-6cosu-1=0\)

\(\Leftrightarrow2\left(4cos^3u-3cosu\right)=1\)

\(\Leftrightarrow2cos3u=1\)

\(\Leftrightarrow cos3u=\dfrac{1}{2}\)

\(\Rightarrow\left[{}\begin{matrix}3u=\dfrac{\pi}{3}+k2\pi\\3u=-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u=\dfrac{\pi}{9}+\dfrac{k2\pi}{3}\\u=-\dfrac{\pi}{9}+\dfrac{k2\pi}{3}\end{matrix}\right.\)

Vậy nghiệm của pt là: \(x=cosu=\left\{cos\left(\dfrac{\pi}{9}\right);cos\left(\dfrac{5\pi}{9}\right);cos\left(\dfrac{7\pi}{9}\right)\right\}\)