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a)\(3x^2-4x=0<=>x(3x-4)=0\)
TH1: x=0
TH2 3x-4=0 <=>x=4/3
KL:.....
b) (x+3)(x−1)+2x(x+3)=0.
<=> (x+3)(x-1+2x)=0
TH1: x+3=0 <=> x=-3
TH2 x-1=0 <=> x=1
KL:.....
c) \(9x^2+6x+1=0. <=>(3x+1)^2=0<=>3x+1=0<=>x=-1/3 \)
KL:......
d) \(x^2−4x=4.<=>(x-2)^2=0<=>x-2=0<=>x=2\)
KL:....
a) \(3x^2-4x=0\)
\(\Leftrightarrow x\left(3x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{3}\end{matrix}\right.\)
b) \(\left(x+3\right)\left(x-1\right)+2x\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{1}{3}\end{matrix}\right.\)
c) \(9x^2+6x+1=0\)
\(\Leftrightarrow\left(3x+1\right)^2=0\)
\(\Leftrightarrow3x+1=0\Leftrightarrow x=-\dfrac{1}{3}\)
d) \(x^2-4x=4\)
\(\Leftrightarrow\left(x-2\right)^2=8\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=2\sqrt{2}\\x-2=-2\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\sqrt{2}+2\\x=-2\sqrt{2}+2\end{matrix}\right.\)
a: =>x^2+4x-4x+1=0
=>x^2+1=0
=>Loại
b: =>2x-6+4=2x+2
=>-2=2(loại)
c: =>2(x+3)-2x-1=1
=>6-1=1
=>5=1(loại)
d =>x+3=0
=>x=-3(loại)
e: =>x^2-3x^2+3x-3x-2=0
=>-2x^2-2=0
=>x^2+1=0
=>Loại
câu a, b, c dễ mà. Bạn áp dụng 7 hằng đẳng thúc là làm đc thoii!!
vd: a) \(\left(9x^2-4\right)\left(x+1\right)=\left(3x+2\right)\left(x^2-1\right)\)
\(\Rightarrow\left(3x-2\right)\left(3x+2\right)\left(x+1\right)=\left(3x+2\right)\left(x-1\right)\left(x+1\right)\)
\(\Rightarrow\left(3x-2\right)\left(3x+2\right)-\left(3x+2\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Rightarrow\left(3x+2\right)\left(x+1\right)[\left(3x-2\right)-\left(x-1\right)]=0\)
\(\Rightarrow\left(3x+2\right)\left(x+1\right)\left(2x-1\right)=0\) (bạn phá ngoặc ra rồi tính là ra bước này)
\(\Leftrightarrow3x+2=0\) hoặc \(x+1=0\) hoặc \(2x-1=0\) ( đến đây bạn chia làm 3 trường hợp r tự tính nhé)
Chúc bạn học tốt!!
d/
\(\Leftrightarrow x^3\left(x+1\right)+\left(x+1\right)=0\)
\(\Leftrightarrow\left(x^3+1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x^3+1=0\end{matrix}\right.\) \(\Rightarrow x=-1\)
e/
\(\Leftrightarrow x^3+x^2-6x-x^2-x+6=0\)
\(\Leftrightarrow x\left(x^2+x-6\right)-\left(x^2+x-6\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x-6\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\\x=-3\end{matrix}\right.\)
a) \(\Rightarrow\dfrac{1}{3}x\left(x-2\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
b) \(\Rightarrow\left(x+5\right)\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=-5\\x=1\end{matrix}\right.\)
c) \(\Rightarrow x\left(x^2-\dfrac{1}{9}\right)=0\Rightarrow x\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{1}{3}\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)
e) \(\Rightarrow\left(x+2\right)\left(x+2-x+2\right)=0\Rightarrow\left(x+2\right).4=0\Rightarrow x=-2\)
f) \(\Rightarrow x\left(2x-3\right)+2\left(2x-3\right)=0\Rightarrow\left(2x-3\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-2\end{matrix}\right.\)
g) \(\Rightarrow2\left(3x-2\right)^2-\left(3x-2\right)\left(3x+2\right)=0\Rightarrow\left(3x-2\right)\left(3x-6\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=2\end{matrix}\right.\)
h) \(\Rightarrow x\left(x+1\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=-2\end{matrix}\right.\)
i) \(\Rightarrow4x\left(x+1\right)+5\left(x+1\right)=0\Rightarrow\left(x+1\right)\left(4x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{5}{4}\end{matrix}\right.\)
chẳng ai giải, thôi mình giải vậy!
a) Đặt \(y=x^2+4x+8\),phương trình có dạng:
\(t^2+3x\cdot t+2x^2=0\)
\(\Leftrightarrow t^2+xt+2xt+2x^2=0\)
\(\Leftrightarrow t\left(t+x\right)+2x\left(t+x\right)=0\)
\(\Leftrightarrow\left(2x+t\right)\left(t+x\right)=0\)
\(\Leftrightarrow\left(2x+x^2+4x+8\right)\left(x^2+4x+8+x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-4\end{cases}}\)vậy tập nghiệm của phương trình là:S={-2;-4}
b) nhân 2 vế của phương trình với 12 ta được:
\(\left(6x+7\right)^2\left(6x+8\right)\left(6x+6\right)=72\)
Đặt y=6x+7, ta được:\(y^2\left(y+1\right)\left(y-1\right)=72\)
giải tiếp ra ta sẽ được S={-2/3;-5/3}
c) \(\left(x-2\right)^4+\left(x-6\right)^4=82\)
S={3;5}
d)s={1}
e) S={1;-2;-1/2}
f) phương trình vô nghiệm
a) \(5x\left(3x-7\right)-15x\left(x-1\right)=3\)
\(\Rightarrow15x^2-35x-15x^2+15x=3\)
\(\Rightarrow-20x=3\)
\(\Rightarrow x=-\dfrac{3}{20}\)
b) \(\left(4x+2\right)\left(6x-3\right)-\left(8x+5\right)\left(3x-4\right)=2\)
\(\Rightarrow24x^2+12x-12x-6-24x^2-15x+24x+20=2\)
\(\Rightarrow9x+14=2\)
\(\Rightarrow9x=-12\)
\(\Rightarrow x=-\dfrac{4}{3}\)
c) \(7x^2-21x=0\)
\(\Rightarrow7x\left(x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}7x=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
d) \(9x^2-6x+1=0\)
\(\Rightarrow\left(3x\right)^2-2.3x+1=0\)
\(\Rightarrow\left(3x-1\right)^2=0\)
\(\Rightarrow3x-1=0\)
\(\Rightarrow3x=1\)
\(\Rightarrow x=\dfrac{1}{3}\)
e) \(16x^2-49=0\)
\(\Rightarrow\left(4x\right)^2-7^2=0\)
\(\Rightarrow\left(4x-7\right)\left(4x+7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}4x-7=0\\4x+7=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}4x=7\\4x=-7\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{4}\\x=-\dfrac{7}{4}\end{matrix}\right.\)
f) \(5x^3-20x=0\)
\(\Rightarrow5x\left(x^2-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}5x=0\\x^2-4=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x^2=4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=2\\x=-2\end{matrix}\right.\)
a) x3 + 3x2 + 3x + 1 = 64
=> (x + 1)3 = 64
=> (x + 1)3 = 43
=> x + 1 = 4 => x = 3
b) x3 + 6x2 + 9x = 4x
=> x3 + 6x2 + 9x - 4x = 0
=> x3 + 6x2 + 5x = 0
=> x3 + 5x2 + x2 + 5x = 0
=> x2(x + 5) + x(x + 5) = 0
=> (x + 5)(x2 + x) = 0
=> (x + 5)x(x + 1) = 0
=> \(\hept{\begin{cases}x=-5\\x=0\\x=-1\end{cases}}\)
c) 4(x - 2)2 = (x + 2)2
=> 4(x2 - 4x + 4) = x2 + 4x + 4
=> 4x2 - 16x + 16 = x2 + 4x + 4
=> 4x2 - 16x + 16 - x2 - 4x - 4 = 0
=> 3x2 - 20x + 12 = 0
=> 3x2 - 18x - 2x + 12 = 0
=> 3x(x - 6) - 2(x - 6) = 0
=> (x - 6)(3x - 2) = 0
=> \(\orbr{\begin{cases}x=6\\x=\frac{2}{3}\end{cases}}\)
d) x4 - 16x2 = 0
=> x2(x2 - 16) = 0
=> \(\orbr{\begin{cases}x^2=0\\x^2=16\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm4\end{cases}}\)
e) x4 - 4x3 + x2 - 4x = 0
=> x4 + x2 - 4x3 - 4x = 0
=> x2(x2 + 1) - 4x(x2 + 1) = 0
=> (x2 - 4x)(x2 + 1) = 0
=> x(x - 4)(x2 + 1) = 0
=> \(\orbr{\begin{cases}x=0\\x=4\end{cases}}\)(vì x2 + 1 \(\ge\)1 > 0 \(\forall\)x)
f) x3 + x = 0 => x(x2 + 1) = 0 => x = 0 (vì x2 + 1 \(\ge1>0\forall\)x)
\(a,x^3+3x^2+3x+1=64\)
\(\left(x+1\right)^3=64\)
\(\left(x+1\right)^3=4^3\)
\(x+1=4\)
\(x=3\)