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a: \(x^3-4x^2-x+4=0\)
=>\(\left(x^3-4x^2\right)-\left(x-4\right)=0\)
=>\(x^2\left(x-4\right)-\left(x-4\right)=0\)
=>\(\left(x-4\right)\left(x^2-1\right)=0\)
=>\(\left[{}\begin{matrix}x-4=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x^2=1\end{matrix}\right.\Leftrightarrow x\in\left\{2;1;-1\right\}\)
b: Sửa đề: \(x^3+3x^2+3x+1=0\)
=>\(x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=0\)
=>\(\left(x+1\right)^3=0\)
=>x+1=0
=>x=-1
c: \(x^3+3x^2-4x-12=0\)
=>\(\left(x^3+3x^2\right)-\left(4x+12\right)=0\)
=>\(x^2\cdot\left(x+3\right)-4\left(x+3\right)=0\)
=>\(\left(x+3\right)\left(x^2-4\right)=0\)
=>\(\left(x+3\right)\left(x-2\right)\left(x+2\right)=0\)
=>\(\left[{}\begin{matrix}x+3=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\\x=-2\end{matrix}\right.\)
d: \(\left(x-2\right)^2-4x+8=0\)
=>\(\left(x-2\right)^2-\left(4x-8\right)=0\)
=>\(\left(x-2\right)^2-4\left(x-2\right)=0\)
=>\(\left(x-2\right)\left(x-2-4\right)=0\)
=>(x-2)(x-6)=0
=>\(\left[{}\begin{matrix}x-2=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)
kho..............wa...................troi................thi......................ret.....................ai..............tich...............ung.....................ho....................minh..................voi................ret............wa
hướng dẫn cách làm-tự làm tiếp nha :)
a) đặt \(k=x^2-4x\), ta có:\(k^2-2k=15\)\(\Rightarrow k^2-2x+1=16\Rightarrow\left(k-1\right)^2=4^2=\left(-4\right)^2\)
b) đặt \(A=x^2-3x\), ta có: \(A^2-2A-8=0\Rightarrow A^2-2A+1=9\Rightarrow\left(A-1\right)^2=3^2=\left(-3\right)^2\)
c)theo đề \(\Leftrightarrow\orbr{\begin{cases}x^2-4x+3=0\\x^2-8x+9=0\end{cases}}\)
\(x^2-4x+3=0\Leftrightarrow x^2-4x+4=1\Leftrightarrow\left(x-2\right)^2=1^2=\left(-1\right)^2\)
\(x^2-8x+9=0\Leftrightarrow x^2-8x+16=7\Leftrightarrow\left(x-4\right)^2=\pm\sqrt{7}^2\)
vt ko chi tiết bn ib là đc rùi, sai tớ làm gì T.T
mà tớ làm mẫu 1 bài thui nha, bài còn lại có cách làm òi. bn tự dựa vô nha
\(\text{Đặt }k=x^2-4x,\text{ta có:}\)
\(\left(x^2-4x\right)^2-2.\left(x^2-4x\right)=15\)
\(\Leftrightarrow k^2-2k=0\)
\(\Leftrightarrow k^2-2k+1=16\)
\(\Leftrightarrow\left(k-1\right)^2=16\)
\(\Leftrightarrow\orbr{\begin{cases}k-1=4\\k-1=-4\end{cases}\Leftrightarrow\orbr{\begin{cases}k=5\\k=-3\end{cases}}}\)
\(\text{Với }k=5,\text{Ta có: }x^2-4x=5\Rightarrow x^2-4x-5=0\Rightarrow x^2-5x+x-5=0\)
\(\Rightarrow x.\left(x-5\right)+\left(x-5\right)=0\Rightarrow\left(x+1\right).\left(x-5\right)=0\Rightarrow\orbr{\begin{cases}x=-1\\x=5\end{cases}}\)
\(\text{Với }k=-3,\text{ta có: }x^2-4x=-3\Rightarrow x^2-4x+3=0\Rightarrow k^2-3x-x+3=0\)
\(\Rightarrow x.\left(x-3\right)-\left(x-3\right)=0\Rightarrow\left(x-1\right).\left(x-3\right)=0\Rightarrow\orbr{\begin{cases}x=1\\x=3\end{cases}}\)
Vậy...
a) \(3x\left(x-2\right)-4x+8=0\)
\(\Leftrightarrow3x\left(x-2\right)-4\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(3x-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\3x-4=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=\frac{4}{3}\end{cases}}\)
b) \(3\left(2x-1\right)^2+2-4x=0\)
\(\Leftrightarrow3\left(2x-1\right)^2-2\left(2x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(6x-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=0\\6x-5=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{5}{6}\end{cases}}\)
a) \(x^2-12x+11\)\(=0\)
\(\Leftrightarrow\left(x-6\right)^2-25=0\)
\(\Leftrightarrow\left(x-6+5\right)\left(x-6-5\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-11=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=11\end{matrix}\right.\)
a)\(x^2-12x+11=0\)
\(x^2-x-11x+11=0\)
\(\left(x^2-x\right)-\left(11x-11\right)=0\)
\(x\left(x-1\right)-11\left(x-1\right)=0\)
\(\left(x-1\right)\left(x-11\right)=0\)
\(=>\left[{}\begin{matrix}x-1=0\\x-11=0\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}x=1\\x=11\end{matrix}\right.\)
b)\(4x^2-4x-3=0\)
\(4x^2-2x+6x-3=0\)
\(2x\left(2x-1\right)+3\left(3x-1\right)=0\)
\(\left(2x-1\right)\left(2x+3\right)=0\)
\(=>\left[{}\begin{matrix}2x-1=0\\2x+3=0\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}x=0,5\\x=-1,5\end{matrix}\right.\)\
c)\(4x^2-12x-7=0\)
\(4x^2-14x+2x-7=0\)
\(2x\left(2x-7\right)+\left(2x-7\right)=0\)
\(\left(2x-7\right)\left(2x+1\right)=0\)
\(=>\left[{}\begin{matrix}2x-7=0\\2x+1=0\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}x=3,5\\x=-0,5\end{matrix}\right.\)
a: \(x^2+12x+36=0\)
=>\(x^2+2\cdot x\cdot6+6^2=0\)
=>\(\left(x+6\right)^2=0\)
=>x+6=0
=>x=-6
b: \(4x^2-4x+1=0\)
=>\(\left(2x\right)^2-2\cdot2x\cdot1+1^2=0\)
=>\(\left(2x-1\right)^2=0\)
=>2x-1=0
=>2x=1
=>x=1/2
c: \(x^3+6x^2+12x+8=0\)
=>\(x^3+3\cdot x^2\cdot2+3\cdot x\cdot2^2+2^3=0\)
=>\(\left(x+2\right)^3=0\)
=>x+2=0
=>x=-2
a) Ta có: \(x^2+3x-10=0\)
\(\Leftrightarrow x^2+5x-2x-10=0\)
\(\Leftrightarrow x\left(x+5\right)-2\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
Vậy: S={-5;2}
b) Ta có: \(3x^2-7x+1=0\)
\(\Leftrightarrow3\left(x^2-\dfrac{7}{3}x+\dfrac{1}{3}\right)=0\)
mà 3>0
nên \(x^2-\dfrac{7}{3}x+\dfrac{1}{3}=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{7}{6}+\dfrac{49}{36}-\dfrac{37}{36}=0\)
\(\Leftrightarrow\left(x-\dfrac{7}{6}\right)^2=\dfrac{37}{36}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{7}{6}=\dfrac{\sqrt{37}}{6}\\x-\dfrac{7}{6}=-\dfrac{\sqrt{37}}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{37}+7}{6}\\x=\dfrac{-\sqrt{37}+7}{6}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{\sqrt{37}+7}{6};\dfrac{-\sqrt{37}+7}{6}\right\}\)
c) Ta có: \(3x^2-7x+8=0\)
\(\Leftrightarrow3\left(x^2-\dfrac{7}{3}x+\dfrac{8}{3}\right)=0\)
mà 3>0
nên \(x^2-\dfrac{7}{3}x+\dfrac{8}{3}=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{7}{6}+\dfrac{49}{36}+\dfrac{47}{36}=0\)
\(\Leftrightarrow\left(x-\dfrac{7}{6}\right)^2=-\dfrac{47}{36}\)(vô lý)
Vậy: \(x\in\varnothing\)
Đặt \(u=x^2-4x\)
Phương trình trở thành \(u^2+2u-8=0\)
\(\Leftrightarrow u^2-2u+4u-8=0\)
\(\Leftrightarrow u\left(u-2\right)+4\left(u-2\right)=0\)
\(\Leftrightarrow\left(u+4\right)\left(u-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}u+4=0\\u-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}u=-4\\u=2\end{cases}}\)
Lúc đó thì \(\orbr{\begin{cases}x^2-4x=-4\\x^2-4x=2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x^2-4x+4=0\\x^2-4x-2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x-2\right)^2=0\\\left(x-2\right)^2=6\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=2\pm\sqrt{6}\end{cases}}\)
Vậy phương trình có 3 nghiệm \(\left\{2;2\pm\sqrt{6}\right\}\)
\(\left(x^2-4x\right)^2+2\left(x^2-4x\right)-8=0\)
\(\Leftrightarrow x^4-8x^3+16x^2+2x^2-8x-8=0\)
\(\Leftrightarrow x^4-8x^3+18x^2-8x-8=0\)
\(\Leftrightarrow x^4-2x^3-6x^3+12x^2+6x^2-12x+4x-8=0\)
\(\Leftrightarrow x^3\left(x-2\right)-6x^2\left(x-2\right)+6x\left(x-2\right)+4\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3-6x^2+6x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x^3-6x^2+6x+4=0\end{cases}}\)
Đoạn này bạn làm từng bước một nhé ! Mình làm tắt thôi :
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=2\pm\sqrt{6}\end{cases}}\)
Vậy tập nghiệm của phương trình là : \(S=\left\{2;2+\sqrt{6};2-\sqrt{6}\right\}\)