1) Tìm x biết
a) x3-3x2+3x-1=0
b) (x-2)3+6(x+1)2-x+12=0
c) x3+6x2+12x+8=0
d) x3-6x2+12x-8=0
e) 8x3-12x3+6x-1=0
f) x3+9x2+27x+27=0 giải giùm mình nha
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a: 49x^2-25=0
=>(7x-5)(7x+5)=0
=>7x-5=0 hoặc 7x+5=0
=>x=5/7 hoặc x=-5/7
b: Đề thiếu vế phải rồi bạn
c: (3x-2)^2-9(x+4)(x-4)=2
=>9x^2-12x+4-9(x^2-16)=2
=>9x^2-12x+4-9x^2+144=2
=>-12x+148=2
=>-12x=-146
=>x=146/12=73/6
d: x^3-6x^2+12x-8=0
=>(x-2)^3=0
=>x-2=0
=>x=2
e: x^3-9x^2+27x-27=0
=>(x-3)^3=0
=>x-3=0
=>x=3
a) \(-25+49x^2=0\)
\(\Leftrightarrow49x^2-25=0\)
\(\Leftrightarrow\left(7x\right)^2-5^2=0\)
\(\Leftrightarrow\left(7x-5\right)\left(7x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}7x-5=0\\7x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}7x=5\\7x=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{7}\\x=-\dfrac{5}{7}\end{matrix}\right.\)
b) \(16x^2-25\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(4x\right)^2-\left[5\left(x-2\right)\right]^2=0\)
\(\Leftrightarrow\left(4x-5x+10\right)\left(4x+5x-10\right)=0\)
\(\Leftrightarrow\left(10-x\right)\left(9x-10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}10-x=0\\9x=10\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=10\\x=\dfrac{10}{9}\end{matrix}\right.\)
c) \(\left(3x-2\right)^2-9\left(x+4\right)\left(x+4\right)=2\)
\(\Leftrightarrow9x^2-12x+4-9\left(x^2+8x+16\right)=2\)
\(\Leftrightarrow9x^2-12x+4-9x^2-72x-144=2\)
\(\Leftrightarrow-84x-140=2\)
\(\Leftrightarrow-84x=142\)
\(\Leftrightarrow x=-\dfrac{142}{84}\)
\(\Leftrightarrow x=-\dfrac{71}{42}\)
d) \(x^3-6x^2+12x-8=0\)
\(\Leftrightarrow x^3-3\cdot2\cdot x^2+3\cdot2^2\cdot x-2^3=0\)
\(\Leftrightarrow\left(x-2\right)^3=0\)
\(\Leftrightarrow x-2=0\)
\(\Leftrightarrow x=2\)
e) \(-27+27x-9x^2+x^3=0\)
\(\Leftrightarrow x^3-9x^2+27x-27=0\)
\(\Leftrightarrow\left(x-3\right)^3=0\)
\(\Leftrightarrow x-3=0\)
\(\Leftrightarrow x=3\)
\(a,PT\Leftrightarrow x^3-6x^2+12x-8-x^3+x+6x^2-18x-10=0\)
\(\Leftrightarrow-5x-18=0\)
\(\Leftrightarrow x=-\dfrac{18}{5}\)
Vậy ...
\(b,PT\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-6+10=0\)
\(\Leftrightarrow12x+6=0\)
\(\Leftrightarrow x=-\dfrac{1}{2}\)
Vậy ...
\(c,PT\Leftrightarrow\left(x+1\right)^3+3^3=0\)
\(\Leftrightarrow\left(x+1+3\right)\left(x^2+2x+1-3x-3+9\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x^2-x+7\right)=0\)
Thấy : \(x^2-\dfrac{2.x.1}{2}+\dfrac{1}{4}+\dfrac{27}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{27}{4}\ge\dfrac{27}{4}>0\)
\(\Rightarrow x+4=0\)
\(\Leftrightarrow x=-4\)
Vậy ...
\(d,PT\Leftrightarrow\left(x-2\right)^3+1^3=0\)
\(\Leftrightarrow\left(x-2+1\right)\left(x^2-4x+4-x+2+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-5x+7\right)=0\)
Thấy : \(x^2-5x+7=x^2-\dfrac{5.x.2}{2}+\dfrac{25}{4}+\dfrac{3}{4}=\left(x-\dfrac{5}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\)
\(\Rightarrow x-1=0\)
\(\Leftrightarrow x=1\)
Vậy ...
1)
a) \(=15x^3-20x^2+10x\)
b) \(=3x^4-x^3+4x^2-9x^3+3x-12x=3x^4-10x^3+4x^2-9x\)
2)
a) \(\Rightarrow x\left(x^2-6x+12\right)=0\)
\(\Rightarrow x=0\)(do \(x^2-6x+12=\left(x^2-6x+\dfrac{36}{4}\right)+3=\left(x-\dfrac{6}{2}\right)^2+3\ge3>0\))
b) \(\Rightarrow\left(x+3\right)^3=0\Rightarrow x=-3\)
(3x²-5x+2)+(3x²+5x)= bao nhiêu ạ
Giúp em vs ạ . Em cảm ơn
a) \(\Rightarrow\left(x-1\right)^3=0\Rightarrow x=1\)
b) \(\Rightarrow\left(x^3-1\right)\left(x^3+1\right)=0\Rightarrow\left(x-1\right)\left(x+1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)(do \(\left\{{}\begin{matrix}x^2-x+1=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\\x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\end{matrix}\right.\))
c) \(\Rightarrow4x\left(x^2-9\right)=0\Rightarrow4x\left(x-3\right)\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
d) \(\Rightarrow\left(x-2\right)^3=0\Rightarrow x=2\)
a) \(x^3-3x^2+3x-1=0\Rightarrow\left(x-1\right)^3=0\Rightarrow x-1=0\)
\(\Rightarrow x=1\)
b) \(x^6-1=0\Rightarrow\left(x^3\right)^2-1=0\Rightarrow\left(x^3-1\right)\left(x^3+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^3-1=0\\x^3+1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
c) \(4x^3-36x=0\Rightarrow4x\left(x^2-36\right)=0\Rightarrow4x\left(x-6\right)\left(x+6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}4x=0\\x-6=0\\x+6=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=6\\x=-6\end{matrix}\right.\)
d) \(x^3-6x^2+12x-8=0\) (đề bài như vậy mới làm đc, nếu là +8 thì mình xin bó tay nhé)
\(\Rightarrow x^3-3\cdot x^2\cdot2+3\cdot x\cdot2^2-2^3=0\)
\(\Rightarrow\left(x-2\right)^3=0\Rightarrow x-2=0\Rightarrow x=2\)
a: \(\Leftrightarrow\left(x-5\right)\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\\x=1\end{matrix}\right.\)
d: \(\Leftrightarrow\left(x+3\right)\left(x^2-4x+5\right)=0\)
\(\Leftrightarrow x+3=0\)
hay x=-3
a) Cách 1: Khai triển HĐT rút gọn được 3 x 2 + 6x + 7 = 0
Vì (3( x 2 + 2x + 1) + 4 < 0 với mọi x nên giải được x ∈ ∅
Cách 2. Chuyển vế đưa về ( x + 3 ) 3 = ( x - 1 ) 3 Û x + 3 = x - 1
Từ đó tìm được x ∈ ∅
b) Đặt x 2 = t với t ≥ 0 ta được t 2 + t - 2 = 0
Giải ra ta được t = 1 (TM) hoặc t = -2 (KTM)
Từ đó tìm được x = ± 1
c) Biến đổi được
d) Biến đổi về dạng x(x - 2) (x - 4) = 0. Tìm được x ∈ {0; 2; 4}
\(a,\Leftrightarrow\left(x-2\right)\left(3x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-2\right)^3=0\Leftrightarrow x-2=0\Leftrightarrow x=2\\ c,\Leftrightarrow\left(4x-3x-3\right)\left(4x+3x+3\right)=0\\ \Leftrightarrow\left(x-3\right)\left(7x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{7}\end{matrix}\right.\\ d,\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
\(a,x^3-3x^2+3x-1=0\)
\(\Leftrightarrow\left(x-1\right)^3=0\)
\(\Rightarrow x-1=0\Rightarrow x=1\)
\(b,\left(x-2\right)^3+6\left(x+1\right)^2-x+12=0\)
\(\Leftrightarrow x^3-6x^2+12x-8+6x^2+12x+6-x+12=0\)\(\Leftrightarrow x^3+23x+10=0\) (1)
Đặt \(t=\dfrac{x}{\dfrac{2\sqrt{69}}{3}}\Leftrightarrow x=\dfrac{2\sqrt{69}}{3}t\)
Khi đó: (1) \(\Leftrightarrow4t^3+3t=-0,2355375386\)
Đặt a= \(\sqrt[3]{-0,2355375386+\sqrt{-0,2355375386^2+1}}\)
Và \(\alpha=\dfrac{1}{2}\left(a-\dfrac{1}{a}\right)\) , ta được:
\(4\alpha^3+3\alpha=-0,2355375386\) , vậy \(t=\alpha\) là nghiệm của pt
Vậy t= \(\dfrac{1}{2}\left(\sqrt[3]{-0,2355375386}+\sqrt{-0,2355375386^2+1}\right)\) \(\left(\sqrt[3]{-0,2355375386-\sqrt{-0,2355375386^2+1}}\right)\)\(=-0,07788262891\)
\(\Rightarrow x=\dfrac{2\sqrt{69}}{3}.t=-0,4312944692\)
\(c,x^3+6x^2+12x+8=0\)
\(\Leftrightarrow\left(x+2\right)^3=0\)
\(\Leftrightarrow x+2=0\Rightarrow x=-2\)
\(d,x^3-6x^2+12x-8=0\)
\(\Leftrightarrow\left(x-2\right)^3=0\)
\(\Rightarrow x-2=0\Rightarrow x=2\)
\(e,8x^3-12x^2+6x-1=0\)
\(\Leftrightarrow\left(2x-1\right)^3=0\)
\(\Rightarrow2x-1=0\Rightarrow x=\dfrac{1}{2}\)
\(f,x^3+9x^2+27x+27=0\)
\(\Leftrightarrow\left(x+3\right)^3=0\)
\(\Rightarrow x+3=0\Rightarrow x=-3\)