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a/ \(\left(x-4\right)^2-36=0\)
<=> \(\left(x-4-6\right)\left(x-4+6\right)=0\)
<=> \(\left(x-10\right)\left(x+2\right)=0\)
<=> \(\orbr{\begin{cases}x-10=0\\x+2=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=10\\x=-2\end{cases}}\)
b/ \(\left(x+8\right)^2=121\)
<=> \(\left(x+8\right)^2-121=0\)
<=> \(\left(x+8-11\right)\left(x+8+11\right)=0\)
<=> \(\left(x-3\right)\left(x+19\right)=0\)
<=> \(\orbr{\begin{cases}x-3=0\\x+19=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=3\\x=-19\end{cases}}\)
d/ \(4x^2-12x+9=0\)
<=> \(\left(2x\right)^2-2.2x.3+3^2=0\)
<=> \(\left(2x-3\right)^2=0\)
<=> \(2x-3=0\)
<=> \(x=\frac{3}{2}\)
Mẫu cho 1 câu
\(B=5-\left|x-3\right|\)
ĐK: \(x>4\Leftrightarrow x-3>0\)
Ta có: \(\left|x-3\right|=x-3\)
\(\Rightarrow B=5-x-3=-x+2\)
+ Nếu \(x-3< 0\Leftrightarrow x< 3\)
Ta có: \(\left|x-3\right|=-\left(x-3\right)=-x+3\)
easy rồi tự thế vô tính tiếp
a: \(A=\left(\dfrac{1}{x-1}+\dfrac{x}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+x+1}{x+1}\right)\cdot\dfrac{\left(x+1\right)^2}{2x+1}\)
\(=\left(\dfrac{1}{x-1}+\dfrac{x}{\left(x-1\right)\left(x+1\right)}\right)\cdot\dfrac{\left(x+1\right)^2}{2x+1}\)
\(=\dfrac{x+1+x}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x+1\right)^2}{2x+1}\)
\(=\dfrac{2x+1}{x-1}\cdot\dfrac{x+1}{2x+1}=\dfrac{x+1}{x-1}\)
b: Thay x=1/2 vào A, ta được:
\(A=\dfrac{\dfrac{1}{2}+1}{\dfrac{1}{2}-1}=\dfrac{3}{2}:\dfrac{-1}{2}=-3\)
c: Để A là số nguyên thì \(x-1+2⋮x-1\)
\(\Leftrightarrow x-1\in\left\{1;-1;2;-2\right\}\)
\(\Leftrightarrow x\in\left\{2;0;3\right\}\)
a) Ta có : \(x^2-6x+10\)
\(=\left(x^2-6x+9\right)+1\)
\(=\left(x-3\right)^2+1\ge1>0\forall x\)
b) Ta có : \(4x-x^2-5\)
\(=-\left(x^2-4x+4\right)-1\)
\(=-\left(x-2\right)^2-1\le-1< 0\forall x\)
Vậy ...
B1:
pt <=> \(\dfrac{3x^2}{10}+\dfrac{2y^2}{15}+\dfrac{z^2}{20}=0\)
<=> x = y = z = 0
B2: Áp dụng bđt Cô-si:
\(\left(x^2+\dfrac{1}{x^2}\right)+\left(y^2+\dfrac{1}{y^2}\right)\ge2+2=4\)
Dấu "=" xảy ra <=> x2 = y2 = 1
\( a)\dfrac{{3{x^4} - 2{x^3} - 2{x^2} + 4x - 8}}{{{x^2} - 2}}\\ = \dfrac{{3{x^4} - 2{x^3} - 6{x^2} + 4{x^2} + 4x - 8}}{{{x^2} - 2}}\\ = \dfrac{{3{x^2}\left( {{x^2} - 2} \right) - 2x\left( {{x^2} - 2} \right) + 4\left( {{x^2} - 2} \right)}}{{{x^2} - 2}}\\ = \dfrac{{\left( {{x^2} - 2} \right)\left( {3{x^2} - 2x + 4} \right)}}{{{x^2} - 2}}\\ = 3{x^2} - 2x + 4 \)
\( b)\dfrac{{2{x^3} - 26x - 24}}{{{x^2} + 4x + 3}}\\ = \dfrac{{2\left( {{x^3} - 13x - 12} \right)}}{{x + 3x + x + 3}}\\ = \dfrac{{2\left( {{x^3} + {x^2} - {x^2} - x - 12x - 12} \right)}}{{x\left( {x + 3} \right) + x + 3}}\\ = \dfrac{{2\left[ {{x^2}\left( {x + 1} \right) - x\left( {x + 1} \right) - 12\left( {x + 1} \right)} \right]}}{{\left( {x + 3} \right)\left( {x + 1} \right)}}\\ = \dfrac{{2\left( {x + 1} \right)\left( {{x^2} - x - 12} \right)}}{{\left( {x + 3} \right)\left( {x + 1} \right)}}\\ = \dfrac{{2\left( {{x^2} + 3x - 4x - 12} \right)}}{{x + 3}}\\ = \dfrac{{2\left[ {x\left( {x + 3} \right) - 4\left( {x + 3} \right)} \right]}}{{x + 3}}\\ = \dfrac{{2\left( {x + 3} \right)\left( {x - 4} \right)}}{{x + 3}}\\ = 2\left( {x - 4} \right)\\ = 2x - 8\)
Bài 1:
\(A=4x^2-28x+49\)
\(=4x^2-16x-12x+48+1\)
\(=4x\left(x-4\right)-12\left(x-4\right)+1\)
\(=\left(4x-12\right)\left(x-4\right)+1\)
Thay x = 4
\(\Leftrightarrow A=1\)
Vậy A= 1 tại x = 4
câu b
\(x^2-x=24\Leftrightarrow x^2-2\dfrac{1}{2}x+\dfrac{1}{4}=24+\dfrac{1}{4}=\dfrac{97}{4}\)
\(\left(x-\dfrac{1}{2}\right)^2=\left(\dfrac{\sqrt{97}}{2}\right)^2\)
\(\left[{}\begin{matrix}x_1=\dfrac{1+\sqrt{97}}{2}\\x_2=\dfrac{1-\sqrt{97}}{2}\end{matrix}\right.\)