Bài 4: a) Rút gọn Q. Q=2x- căn(x^2-2x+1) . b) tìm x để Q=7
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a) \(Q^2=2x-\sqrt{\left(x^2+2x+1\right)}\)
\(Q^2=2x-\sqrt{\left(x+1\right)^2}\)
\(Q^2=2x-x-1\)
\(Q^2=x-1\)
\(Q=\sqrt{x-1}\)
b) Để Q=9 thì \(\sqrt{x-1}=9\) => \(x=82\)
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điều kiện \(x\ge0\)và x khác 1/4
Q= \(\frac{3\sqrt{x}+2}{2\sqrt{x}-1}+\frac{\sqrt{x}-1}{\sqrt{x}+4}-\frac{x-6\sqrt{x}+5}{2x+7\sqrt{x}-4}=\frac{3x+14\sqrt{x}+8+2x-3\sqrt{x}+1-x+6\sqrt{x}-5}{2x+7\sqrt{x}-4}\)
=\(\frac{4x+17\sqrt{x}+4}{2x+7\sqrt{x}-4}\)
đề Q>1/2 thì \(\frac{4x+17\sqrt{x}+4}{2x+7\sqrt{x}-4}>\frac{1}{2}\)
<=> \(8x+34\sqrt{x}+8>2x+7\sqrt{x}-4\)<=> \(6x+27\sqrt{x}+12>0\) với mọi x>=0
vậy Q>1/2 khi x>=0 và x khác 1/4
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Bài 2:
3x + 2(5 - x) = 0
<=> 3x + 10 - 2x = 0
<=> x + 10 = 0
<=> x = 0 - 10
<=> x = -10
=> x = -10
Bài 3:
6(3q + 4q) - 8(5p - q) + (p - q)
= 6.3p + 6.4q - 8.5p - (-8).q + p - q
= 18p + 24q - 40p + 8q + p - q
= (18p - 40p + p) + (24q + 8q - q)
= -21p + 31q
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1.
a. ĐKXĐ : x lớn hơn hoặc bằng 1/2
b. A\(\sqrt{2}\)= \(\sqrt{2x+2\sqrt{2x-1}}-\sqrt{2x-2\sqrt{2x-1}}\)
= \(\sqrt{2x-1+1+2\sqrt{2x-1}}-\sqrt{2x-1+1-2\sqrt{2x-1}}\)
=\(\sqrt{\left(\sqrt{2x-1}+1\right)^2}-\sqrt{\left(\sqrt{2x-1}-1\right)^2}\)
= \(\sqrt{2x-1}+1-\left|\sqrt{2x-1}-1\right|\)
Nếu \(x\ge1thìA\sqrt{2}=\sqrt{2x-1}+1-\left(\sqrt{2x-1}-1\right)=2\)
\(\Rightarrow A=2\)
Nếu 1/2 \(\le x< 1thìA\sqrt{2}=\sqrt{2x-1}+1-\left(1-\sqrt{2x-1}\right)=2\sqrt{2x-1}\)
Do đó : A= \(\sqrt{4x-2}\)
Vậy ............
2.
a. \(x\ge2\)hoặc x<0
b. A= \(2\sqrt{x^2-2x}\)
c. A<2 \(\Leftrightarrow\)\(2\sqrt{x^2-2x}< 2\Leftrightarrow\sqrt{x^2-2x}< 1\Leftrightarrow x^2-2x< 1\Leftrightarrow\left(x-1\right)^2< 2\)
\(-\sqrt{2}< x-1< \sqrt{2}\Leftrightarrow1-\sqrt{2}< x< 1+\sqrt{2}\)
Kết hợp vs đk câu a , ta đc : \(1-\sqrt{2}< x< 0và2\le x< 1+\sqrt{2}\)
Vậy...........
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a: ĐKXĐ: x<>2; x<>3
\(Q=\dfrac{2x-9-x^2+9+2x^2-4x+x-2}{\left(x-3\right)\left(x-2\right)}\)
\(=\dfrac{x^2-x-2}{\left(x-3\right)\left(x-2\right)}=\dfrac{x+1}{x-3}\)
b: Để P<1 thì P-1<0
=>\(\dfrac{x+1-x+3}{x-3}< 0\)
=>x-3<0
=>x<3
a) ĐKXĐ: \(x^2-2x+1\ge0\Rightarrow\left(x-1\right)^2\ge0\left(luônđúng\right)\)
\(Q=2x-\sqrt{x^2-2x+1}=2x-\sqrt{\left(x-1\right)^2}=2x-x+1=x+1\)
b) \(Q=x+1\\ \Rightarrow7=x+1\\ \Rightarrow x=6\)
a: Ta có: \(Q=2x-\sqrt{x^2-2x+1}\)
\(=2x-\left|x-1\right|\)
\(=\left[{}\begin{matrix}2x-x+1=x+1\left(x\ge1\right)\\2x+x-1=3x-1\left(x< 1\right)\end{matrix}\right.\)
b: Ta có: Q=7
nên \(\left[{}\begin{matrix}x+1=7\left(x\ge1\right)\\3x-1=7\left(x< 1\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\left(nhận\right)\\x=\dfrac{8}{3}\left(loại\right)\end{matrix}\right.\)