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a)\(x^2+10x=24\)
\(\Leftrightarrow x^2+10x-24=0\)
\(\Leftrightarrow x^2-2x+12x-24=0\)
\(\Leftrightarrow x\left(x-2\right)+12\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+12\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-2=0\\x+12=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=-12\end{array}\right.\)
b)\(4x^2+4x=24\)
\(\Leftrightarrow4x^2+4x-24=0\)
\(\Leftrightarrow4\left(x^2+x-6\right)=0\)
\(\Leftrightarrow x^2+x-6=0\)
\(\Leftrightarrow x^2+3x-2x-6=0\)
\(\Leftrightarrow x\left(x+3\right)-2\left(x+3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-2=0\\x+3=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=-3\end{array}\right.\)
c)\(4x^2-4x=48\)
\(\Leftrightarrow4x^2-4x-48=0\)
\(\Leftrightarrow4\left(x^2-x-12\right)=0\)
\(\Leftrightarrow x^2-x-12=0\)
\(\Leftrightarrow x^2+3x-4x-12=0\)
\(\Leftrightarrow x\left(x+3\right)-4\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x+3=0\\x-4=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-3\\x=4\end{array}\right.\)
\(a,x^2+10x=24\)
\(\Leftrightarrow x^2+10x-24=0\)
\(\Leftrightarrow x^2-2x+12x-24=0\)
\(\Leftrightarrow x\left(x-2\right)+12\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+12\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-2=0\\x+12=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=-12\end{array}\right.\)
\(\text{Vậy x=2 hoặc x=-12 }\)
\(b,4x^2+4x=24\)
\(\Leftrightarrow4x^2+4x-24=0\)
\(\Leftrightarrow4x^2-8x+12x-24=0\)
\(=4x\left(x-2\right)+12\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(4x+12\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-2=0\\4x+12=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=-3\end{array}\right.\)
Vậy hoặc \(\text{Vậy x=2 hoặc x=-3 }\)
\(c,4x^2-4x=48\)
\(\Leftrightarrow4x^2-4x-48=0\)
\(\Leftrightarrow\left[\left(2x\right)^2-2.2x+1^2\right]-1^2-48=0\)
\(\Leftrightarrow\left(2x-1\right)^2-49=0\)
\(\Leftrightarrow\left(2x-1\right)^2-7^2=0\)
\(\Leftrightarrow\left(2x-1-7\right)\left(2x-1+7\right)=0\)
\(\Leftrightarrow\left(2x-8\right)\left(2x+6\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}2x-8=0\\2x+6=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=4\\x=-3\end{array}\right.\)
\(\text{Vậy x=4 hoặc x=-3
}\)
f: Ta có: \(16x^2-9\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(4x-3x-3\right)\left(4x+3x+3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(7x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{7}\end{matrix}\right.\)
ta có:\(x^3+x^2+2x^2+2x+2x+2=0\)0
\(\Leftrightarrow x^2\left(x+1\right)+2x\left(x+1\right)+2\left(x+1\right)=0\)
\(\Leftrightarrow\left(x^2+2x+2\right)\left(x+1\right)=0\)
Do \(x^2+2x+2\ne0\)
\(\Rightarrow x+1=0\)
\(\Rightarrow x=-1\)
vậy phương trình trên có tập nghiệm là :S=(-1)
x2 - 4x + 2 = ( x2 - 4x + 4 ) - 2 = ( x - 2 )2 - 2 ≥ -2 ∀ x
Dấu "=" xảy ra <=> x = 2 . Vậy GTNN của bthuc = -2
x^2 - 4x + 2
= x^2 - 4x + 4 - 2
= ( x - 2 ) ^2 - 2
\(\left(x-2\right)^2\ge0\forall x\)
\(\left(x-2\right)^2-2\ge-2\)
Dấu = xảy ra khi và chỉ khi
x - 2 = 0
x = 0 + 2
x = 2
vậy min = -2 khi và chỉ khi x = 2
\(a,A=\left(x^2-x\right)\left(x^2-x-12\right)\\ A=\left(x^2-x\right)^2-12\left(x^2-x\right)\\ A=\left(x^2-x\right)^2-12\left(x^2-x\right)+36-36\\ A=\left(x^2-x+6\right)^2-36\ge-36\\ A_{min}=-36\Leftrightarrow x^2-x+6=0\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\\ b,B=4x^4+4x^3+5x^2+4x+3\\ B=\left(4x^4+4x^3+x^2\right)+\left(x^2+4x+4\right)-1\\ B=x^2\left(2x+1\right)^2+\left(x+2\right)^2-1\ge-1\\ B_{min}=-1\Leftrightarrow\left\{{}\begin{matrix}x\left(2x+1\right)=0\\x+2=0\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
Vậy dấu \("="\) không xảy ra
\(5x^2+5y^2+8xy-2x+2y+2=0\)
\(\left(4x^2+8xy+4y^2\right)+\left(x^2-2x+1\right)+\left(y^2+2y+1\right)=0\)
\(\left(2x+2y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)
\(4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)
Vì \(\left(x+y\right)^2\ge0;\left(x-1\right)^2\ge0;\left(y+1\right)^2\ge0\)
Để \(4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)
\(\Leftrightarrow x+y=0\)
\(\Leftrightarrow y+1=0\Rightarrow y=-1\)
\(\Leftrightarrow x-1=0\Rightarrow x=1\)
Vậy \(x=1; y=-1\)
Bài làm:
Ta có: \(4x^2-4x-3=0\)
\(\Leftrightarrow\left(4x^2-4x+1\right)-4=0\)
\(\Leftrightarrow\left(2x-1\right)^2-2^2=0\)
\(\Leftrightarrow\left(2x-3\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-3=0\\2x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=-\frac{1}{2}\end{cases}}\)
Ta có : \(4x^2-4x-3=0\)
\(\Leftrightarrow\left(4x^2-4x+1\right)-4=0\)
\(\Leftrightarrow\left(2x-1\right)^2=4\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=2\\2x-1=-2\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{3}{2}\\x=-\frac{1}{2}\end{cases}}\)
Vậy \(x\in\left\{\frac{3}{2};-\frac{1}{2}\right\}\)