nghiệm lẻ đề có sai không vậy

1)(x²-4x+16)(x+4)-x(x+1)(x+2)+3x²=0

\(\Rightarrow\)(x³+64)-x(x²+2x+x+2)+3x²=0

\(\Rightarrow\)x³+64-x³-2x²-x²-2x+3x²=0

\(\Rightarrow\)-2x+64=0

\(\Rightarrow\)-2x=-64

\(\Rightarrow\)x=\(\dfrac{-64}{-2}\)

\(\Rightarrow x=32\)

2)(8x+2)(1-3x)+(6x-1)(4x-10)=-50

\(\Rightarrow\)8x-24x²+2-6x+24x²-60x-4x+10=50

\(\Rightarrow\)-62x+12=50

\(\Rightarrow\)-62x=50-12

\(\Rightarrow\)-62x=38

\(\Rightarrow\)x=\(-\dfrac{38}{62}=-\dfrac{19}{31}\)

đằng giữa 2 căn là dấu cộng nha ~

a) Thực hiện rút gọn VT = -2x – 64

Giải phương trình -2x – 64 = 0 thu được x = -32.

b) Thực hiện rút gọn VT = -62 x +12

Giải phương trình -62x + 12 = -50 thu được x = 1.

1) Ta có: \(5\left(x-3\right)\left(x-7\right)-\left(5x+1\right)\left(x-2\right)=-8\)

\(\Leftrightarrow5\left(x^2-10x+21\right)-\left(5x^2-10x+x-2\right)=-8\)

\(\Leftrightarrow5x^2-50x+105-5x^2+9x+2+8=0\)

\(\Leftrightarrow-41x=-115\)

hay \(x=\dfrac{115}{41}\)

2) Ta có: \(x\left(x+1\right)\left(x+2\right)-\left(x+4\right)\left(3x-5\right)=84-5x\)

\(\Leftrightarrow x\left(x^2+3x+2\right)-\left(3x^2+7x-20\right)=84-5x\)

\(\Leftrightarrow x^3+3x^2+2x-3x^2-7x+20-84+5x=0\)

\(\Leftrightarrow x^3=64\)

hay x=4

3) Ta có: \(\left(9x^2-5\right)\left(x+3\right)-3x^2\left(3x+9\right)=\left(x-5\right)\left(x+4\right)-x\left(x-11\right)\)

\(\Leftrightarrow9x^3+27x^2-5x-15-9x^3-27x^2=x^2-x-20-x^2+11x\)

\(\Leftrightarrow-5x-15=10x-20\)

\(\Leftrightarrow-5x-10x=-20+15\)

\(\Leftrightarrow x=\dfrac{-5}{-15}=\dfrac{1}{3}\)

\(\left|5\left(2x+3\right)\right|+\left|2\left(2x+3\right)\right|+\left|2x+3\right|=16\)

\(=8\left(2x+3\right)=16\)

\(\Rightarrow2x+3=2\)

\(\Rightarrow x=-\frac{1}{2}\)

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: \(x=cosu=\left\{cos\left(\dfrac{\pi}{9}\right);cos\left(\dfrac{5\pi}{9}\right);cos\left(\dfrac{7\pi}{9}\right)\right\}\)