(4x²-8x+3)²-(3x+2)²=0
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a: Ta có: \(x^2+3x+4=0\)
\(\text{Δ}=3^2-4\cdot1\cdot4=9-16=-7< 0\)
Do đó: Phương trình vô nghiệm
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a) \(\left(x+\frac{1}{x}\right)^2+2\left(x+\frac{1}{x}\right)-8=0\)
\(\Leftrightarrow x^2+2x+\frac{1}{x^2}+\frac{2}{x}-6=0\)
\(\Leftrightarrow x^2x^2+2xx^2+\frac{1}{x^2}x^2+\frac{2}{x^2}x^2-6x^2=0.x^2\)
\(\Leftrightarrow x^4+2x^3+1+2x-6x^2=0\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=\sqrt{3}\end{cases}}\)
b) \(x^3-8x^2-8x+1=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-9x+1\right)=0\)
\(\Rightarrow x=-1\)
c) \(x^4-3x^3+4x^2-3x+1=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(x^2-x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}\)
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\(4x^2+4x-3=0\)
\(\left[\left(2x\right)^2+2.2x.1+1\right]-4=0\)
\(\left(2x+1\right)^2-2^2=0\)
\(\left(2x+1-2\right).\left(2x+1+2\right)=0\)
\(\left(2x-1\right).\left(2x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-1=0\\2x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{3}{2}\end{cases}}}\)
Vậy \(\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{3}{2}\end{cases}}\)
\(x^4-3x^3-x+3=0\)
\(x^3.\left(x-3\right)-\left(x-3\right)=0\)
\(\left(x-3\right).\left(x^3-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x^3-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}}\)
Vậy \(\orbr{\begin{cases}x=3\\x=1\end{cases}}\)
\(x^2.\left(x-1\right)-4x^2+8x-4=0\)
\(x^2.\left(x-1\right)-\left[\left(2x\right)^2-2.2x.2+2^2\right]=0\)
\(x^2.\left(x-1\right)-\left(2x-2\right)^2=0\)
\(x^2.\left(x-1\right)-4.\left(x-1\right)^2=0\)
\(\left(x-1\right).\left[x^2-4.\left(x-1\right)\right]=0\)
\(\left(x-1\right).\left[x^2-2.x.2+2^2\right]=0\)
\(\left(x-1\right).\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}}\)
Vậy \(\begin{cases}x=1\\x=2\end{cases}\)
Tham khảo nhé~
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a) 3x2 + 8x + 4 = 0
=> 3x2 + 6x + 2x + 4 = 0
=> 3x(x + 2) + 2(x + 2) = 0
=> (3x + 2)(x + 2) = 0
=> \(\orbr{\begin{cases}3x+2=0\\x+2=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{2}{3}\\x=-1\end{cases}}\)
b) 4x2 - 4x - 3 = 0
=> 4x2 - 6x + 2x - 3 = 0
=> 2x(2x - 3) + (2x - 3) = 0
=> (2x + 1)(2x - 3) = 0
=> \(\orbr{\begin{cases}2x+1=0\\2x-3=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{3}{2}\end{cases}}\)
\(a,3x^2+8x+4=0\)
\(\Rightarrow3x^2+6x+2x+4=0\)
\(\Rightarrow3x\left(x+2\right)+2\left(x+2\right)=0\)
\(\Rightarrow\left(3x+2\right)\left(x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x+2=0\\x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{-2}{3}\\x=-2\end{cases}}}\)
Vậy....
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\(\left|2+3x\right|=\left|4x-3\right|\)
\(\Leftrightarrow\orbr{\begin{cases}2+3x=4x-3\\2+3x=3-4x\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\\x=\frac{1}{7}\end{cases}}\)
Vậy \(x\in\left\{\frac{1}{7};5\right\}\)
\(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=1-4x\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}\)
Vậy \(x\in\left\{\frac{1}{11};\frac{3}{5}\right\}\)
\(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
\(\Leftrightarrow\left|\frac{5}{4}x-\frac{7}{2}\right|=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
Giải tiếp tương tự
Sau đó giải tiếp câu còn lại
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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...
Ta có: \(\left(4x^2-8x+3\right)^2-\left(3x+2\right)^2=0\)
\(\Rightarrow\left(4x^2-8x+3\right)^2=\left(3x+2\right)^2\)
\(\Rightarrow4x^2-8x+3=3x+2\)
\(\Rightarrow4x^2-8x-3x=2-3\)
\(\Rightarrow x\left(4x-8-3\right)=-1\)
\(\Rightarrow5x-11=-1\)
\(\Rightarrow5x=-1 +11\)
\(\Rightarrow5x=10\)
\(\Rightarrow x=2\)