Giúp mình b2 b3 với ạ
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TP
1
5 tháng 3 2023
2.
a.
\(P=\dfrac{1}{x+5}+\dfrac{2}{x-5}-\dfrac{2\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{1}{x+5}+\dfrac{2}{x-5}-\dfrac{2}{x-5}=\dfrac{1}{x+5}\)
b.
\(P=-3\Rightarrow\dfrac{1}{x+5}=-3\Rightarrow x+5=-\dfrac{1}{3}\)
\(\Rightarrow x=-\dfrac{16}{3}\)
Thay vào bấm máy ta được \(Q=529\)
3.
a. \(P=\dfrac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{x+3}{\left(x-3\right)\left(x+3\right)}+\dfrac{18}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{3\left(x-3\right)+x+3+18}{\left(x-3\right)\left(x+3\right)}=\dfrac{4x+12}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{4\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{4}{x-3}\)
b.
\(P=4\Rightarrow\dfrac{4}{x-3}=4\Rightarrow x-3=1\)
\(\Rightarrow x=4\)
NL
Nguyễn Lê Phước Thịnh
CTVHS
4 giờ trước (16:31)
Bài 5:
1: ΔABC vuông tại B
=>\(BA^2+BC^2=AC^2\)
=>\(AC^2=8^2+6^2=64+36=100=10^2\)
=>AC=10(cm)
Bài 1:
a: \(4x^2-4-\left(2x-1\right)\left(3x+4\right)=0\)
=>\(4x^2-4-\left(6x^2+8x-3x-4\right)=0\)
=>\(4x^2-4-\left(6x^2+5x-4\right)=0\)
=>\(4x^2-4-6x^2-5x+4=0\)
=>\(-2x^2-5x=0\)
=>-x(2x+5)=0
=>x(2x+5)=0
=>x∈{0;-5/2}
b: \(\frac{2x+1}{5}-\frac12=\frac{x-2}{6}-\frac{3-x}{4}\)
=>\(\frac{12\left(2x+1\right)}{60}-\frac{30}{60}=\frac{10\left(x-2\right)}{60}-\frac{15\left(3-x\right)}{60}\)
=>12(2x+1)-30=10(x-2)-15(3-x)
=>24x+12-30=10x-20-45+15x
=>24x-18=25x-65
=>24x-25x=-65+18
=>-x=-47
=>x=47
c: ĐKXĐ: x<>3; x<>-3
\(\frac{2}{x+3}=\frac{x-5}{x^2-9}-\frac{5}{3-x}\)
=>\(\frac{2}{x+3}-\frac{x-5}{\left(x-3\right)\left(x+3\right)}-\frac{5}{x-3}=0\)
=>\(\frac{2\left(x-3\right)-\left(x-5\right)-5\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=0\)
=>2(x-3)-(x-5)-5(x+3)=0
=>2x-6-x+5-5x-15=0
=>-4x-16=0
=>x=-4(nhận)
d: |x+2|=2x-3
=>\(\begin{cases}2x-3\ge0\\ \left(2x-3\right)^2=\left(x+2\right)^2\end{cases}\Rightarrow\begin{cases}x\ge\frac32\\ \left(2x-3-x-2\right)\left(2x-3+x+2\right)=0\end{cases}\)
=>\(\begin{cases}x\ge\frac32\\ \left(x-5\right)\left(3x-1\right)=0\end{cases}\)
=>x=5
NL
Nguyễn Lê Phước Thịnh
CTVHS
25 tháng 8 2021
Bài 2:
a: Để (d)//(d') thì \(m=2m+1\)
\(\Leftrightarrow-m=1\)
hay m=-1
c: Để (d) cắt (d') thì \(m\ne2m+1\)
hay \(m\ne-1\)
DT
1
3 tháng 9 2018
a+b=-2
=>(a+b)2=4
=>a2+2ab+b2=4 mà a2+b2=29
=>2ab=-25=>ab=-12,5
=>a2-ab+b2=29-(-12,5)=41,5.
=>(a+b)(a2-ab+b2)=-2.41,5=-83
hay a3+b3=-83
\(2,\\ 1,=20\sqrt{3}+20\sqrt{3}+\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}=40\sqrt{3}+\sqrt{3}=41\sqrt{3}\\ 2,A=\dfrac{2\sqrt{x}-9-x+9+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ A=\dfrac{2\sqrt{x}-x+2x-3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\\ c,A< 1\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}-3}-1< 0\\ \Leftrightarrow\dfrac{4}{\sqrt{x}-3}< 0\Leftrightarrow\sqrt{x}-3< 0\left(4>0\right)\\ \Leftrightarrow x< 9\Leftrightarrow0\le x< 9\)
\(3,\\ 1,A=\sqrt{2}-1-\dfrac{\sqrt{2}\left(2-\sqrt{5}\right)}{2-\sqrt{5}}=\sqrt{2}-1-\sqrt{2}=-1\\ 2,\\ a,P=\dfrac{\sqrt{x}+2-\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\left(\sqrt{x}+2\right)^2}{4}\left(x\ge0;x\ne4\right)\\ P=\dfrac{4\left(\sqrt{x}+2\right)}{4\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}\\ b,P< 1\Leftrightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}-2}-1< 0\\ \Leftrightarrow\dfrac{4}{\sqrt{x}-2}< 0\Leftrightarrow\sqrt{x}-2< 0\left(4>0\right)\\ \Leftrightarrow x< 4\Leftrightarrow0\le x< 4\)