Rút Gọn Biểu Thức
\(A=\left|x-1\right|-\left(2x-5\right)\)
Với giá trị của x thì A=0
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DL
5 tháng 10 2019
a, ĐKXĐ: \(\hept{\begin{cases}x^3+1\ne0\\x^9+x^7-3x^2-3\ne0\\x^2+1\ne0\end{cases}}\)
b, \(Q=\left[\left(x^4-x+\frac{x-3}{x^3+1}\right).\frac{\left(x^3-2x^2+2x-1\right)\left(x+1\right)}{x^9+x^7-3x^2-3}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)
\(Q=\left[\frac{\left(x^3+1\right)\left(x^4-x\right)+x-3}{\left(x+1\right)\left(x^2-x+1\right)}.\frac{\left(x-1\right)\left(x+1\right)\left(x^2-x+1\right)}{\left(x^7-3\right)\left(x^2+1\right)}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)
\(Q=\left[\left(x^7-3\right).\frac{\left(x-1\right)}{\left(x^7-3\right)\left(x^2+1\right)}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)
\(Q=\frac{x-1+x^2+1-2x-12}{x^2+1}\)
\(Q=\frac{\left(x-4\right)\left(x+3\right)}{x^2+1}\)
17 tháng 7 2019
A = 5(x + 3)(x - 3) + (2x + 3)3 + (x - 6)2
A = 5(x + 3)(x - 3) + 4x2 + 12x + 9 + x2 - 12x + 36
A = 5x2 - 45x + 4x2 + 12x + 9 + x2 - 12x + 36
A = 10x2 (1)
Thay x = -1/5 vào (1), ta có:
A = 10x2 = 10.(-1/5)2 = 2/5
A = 2/5
Vậy:...
12 tháng 11 2021
a: \(P=\dfrac{x\left(x+2\right)}{2\left(x+5\right)}+\dfrac{x-5}{x}-\dfrac{5x-50}{2x\left(x+5\right)}\)
\(=\dfrac{x^3+2x^2+2x^2-50-5x+50}{2x\left(x+5\right)}\)
\(=\dfrac{x^3+4x^2-5x}{2x\left(x+5\right)}\)
\(=\dfrac{x\left(x+5\right)\left(x-1\right)}{2x\left(x+5\right)}=\dfrac{x-1}{2}\)
TN
18 tháng 1 2020
Ta có : \(f\left(x\right)=\left|x-1\right|-\left(2x-5\right)\)
Xét 2 TH:
+) Nếu \(\left|x-1\right|=x-1\)
=> \(f\left(x\right)=x-1-2x+5\)
=> \(f\left(x\right)=4-x\)
+) Nếu \(\left|x-1\right|=1-x\)
=> \(f\left(x\right)=1-x-2x+5\)
=> \(f\left(x\right)=6-3x\)
Vậy...
b) \(f\left(5\right)=\left|5-1\right|-\left(2.5-5\right)\)
=> \(f\left(5\right)=4-2=2\)
Vậy...
c) \(f\left(x\right)=0\)
=> \(\left|x-1\right|-\left(2x-5\right)=0\)
=> \(\left|x-1\right|=2x-5\)
Vì \(\left|x-1\right|\ge0\forall x\)
=> \(2x-5\ge0\)
=> \(x\ge\frac{5}{2}\)
=> \(x-1\ge\frac{5}{2}-1=\frac{3}{2}>0\)
=> \(\left|x-1\right|=x-1\)
=> \(x-1-2x+5=0\)
=> \(4-x=0\)
=> \(x=4\)
Y
25 tháng 7 2023
\(a,P=\dfrac{3\left(x+2\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\left(dk:x\ge0,x\ne1\right)\)
\(=\dfrac{3\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{3\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{3\sqrt{x}-\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{2\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{2\left(\sqrt{x}-1\right)}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{2\left(\sqrt{x}+2\right)-\left(\sqrt{x}+1\right)}{\sqrt{x}+2}\\ =\dfrac{2\sqrt{x}+4-\sqrt{x}-1}{\sqrt{x}+2}\\ =\dfrac{\sqrt{x}+3}{\sqrt{x}+2}\)
\(b,x=6-2\sqrt{5}=\left(\sqrt{5}-1\right)^2\)
\(\Rightarrow P=\dfrac{\sqrt{\left(\sqrt{5}-1\right)^2}+3}{\sqrt{\left(\sqrt{5}-1\right)^2}+2}=\dfrac{\left|\sqrt{5}-1\right|+3}{\left|\sqrt{5}-1\right|+2}=\dfrac{\sqrt{5}-1+3}{\sqrt{5}-1+2}=\dfrac{\sqrt{5}+2}{\sqrt{5}+1}\)
TH1 : A = |x-1| - (2x-5)
A = x-1 -2x+5
A = -x +4
TH2 : A = |x-1| - (2x-5)
A = -x+1- 2x +5
A = -3x +6
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