a, \(\left|x^2+2x+1\right|=\left|x^2-x+2\right|\)

\(\Leftrightarrow x^2+2x+1=\pm\left(x^2-x+2\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+2x+1=x^2-x+2\\x^2+2x+1=-x^2+x-2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+2x+1-x^2+x-2=0\\x^2+2x+1+x^2-x+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\left(1\right)\\2x^2-x+3=0\left(2\right)\end{matrix}\right.\)

Giải (1) có : \(3x-1=0\) \(\Leftrightarrow x=\dfrac{1}{3}\)

Giải (2) có : \(2x^2-x+3=0\Rightarrow pt\) \(\text{vô nghiệm }\)

Vậy nghiệm của pt là \(\dfrac{1}{3}\)

b, \(\left|x^2+2x+3\right|=2x+3\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x+3\ge0\text{(1) }\\\left[{}\begin{matrix}x^2+2x+3=2x+3\text{(2) }\\x^2+2x+3=-\left(2x+3\right)\text{(3) }\end{matrix}\right.\end{matrix}\right.\)

Giải (1) có : \(2x+3\ge0\Rightarrow x\ge\dfrac{3}{2}\)

Giải (2) ta có : \(x^2+2x+3=2x+3\)

\(\Leftrightarrow x^2=0\Rightarrow x=0\left(TM\right)\)

Giải (3) ta có : \(x^2+2x+3=-2x-3\)

\(\Leftrightarrow x^2+4x+6=0\Rightarrow pt\) vô nghiêm

Vậy pt có nghiệm là x = 0

Đăng cái lý thuyết làm gì thế

Hoặc bạn có thể giải cách này cho nhanh .

\(x^2=2+\sqrt{2-x}\) ĐK : \(x\le2\)

\(\Leftrightarrow x^2-4=\sqrt{2-x}-2\)

\(\Leftrightarrow\left(x+2\right).\left(x-2\right)=\dfrac{-\left(x+2\right)}{\sqrt{2-x}+2}\)

\(\Leftrightarrow\left(x+2\right).\left(x-2+\dfrac{1}{\sqrt{2-x}+2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\left(TM\right)\\-x+2=\dfrac{1}{\sqrt{2-x}+2}\left(\text{vô nghiệm }\right)\end{matrix}\right.\)

\(x^2=2+\sqrt{2-x}\) ĐK : \(x\le2\)

Đặt \(\sqrt{2-x}=y\ge0\rightarrow y^2=2-x\) . Ta có hệ :

\(\left\{{}\begin{matrix}x^2=y+2\\y^2=2-x\end{matrix}\right.\Rightarrow x^2-y^2=x+y\)

\(\Leftrightarrow\left(x+y\right).\left(x-y-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-y\\x=y+1\end{matrix}\right.\)

Với x = -y ta có pt : \(\sqrt{2+y}=y\)

\(\Leftrightarrow2+y=y^2\)

\(\Leftrightarrow y^2-y-2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y=-1\left(l\right)\\y=2\left(tm\right)\Rightarrow x=-2\left(tm\right)\end{matrix}\right.\)

Với x = y + 1 ta có : \(\sqrt{1-y}=y\)

\(\Leftrightarrow\left\{{}\begin{matrix}y\ge0\\y^2+y-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y\ge0\\\left[{}\begin{matrix}y=\dfrac{-1-\sqrt{5}}{2}\left(l\right)\\y=\dfrac{-1+\sqrt{5}}{2}\Rightarrow x=\dfrac{1+\sqrt{5}}{2}\end{matrix}\right.\end{matrix}\right.\)

\(A=1+3^2+3^4+3^6+....+3^{100}\)

\(3^2A=3^2+3^4+3^6+....+3^{100}+3^{102}\)

\(9A-A=\left(3^2+3^4+3^6+...+3^{100}+3^{102}\right)-\left(1+3^2+3^4+3^6+...+3^{100}\right)\)

\(8A=\dfrac{3^{102}-1}{8}\)

\(4x^2+\sqrt{3x+1}=13x-5\) ĐK : \(x\ge-\dfrac{1}{3}\)

\(\Leftrightarrow4x^2-13x+5=\sqrt{3x+1}\)

\(\Leftrightarrow\left(2x-3\right)^2=-\sqrt{3x+1}+x+4\)

Đặt \(\sqrt{3x+1}=\left(2y-3\right)\) (ĐK : \(y\le\dfrac{3}{2}\))

\(\Leftrightarrow3x+1=\left(2y-3\right)^2\)

Ta có hệ : \(\left\{{}\begin{matrix}3x+1=\left(2y-3\right)^2\\\left(2x-3\right)^2=2y-3+x+4\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-3\right)^2=2y-3+x+4\\\left(2y-3\right)^2=3x+1\end{matrix}\right.\)

\(\Rightarrow\left(2x-3\right)^2-\left(2y-3\right)^2=2y-2x\)

\(\Leftrightarrow2.\left(x-y\right).\left(2x+2y-6\right)=-2.\left(x-y\right)\)

\(\Leftrightarrow\left(x-y\right).\left(2x+2y-6+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=y\\2x+2y-5=0\end{matrix}\right.\)

Với x = y

\(\sqrt{3x+1}=3-2x\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{3}{2}\\3x+1=4x^2-12x+9\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{3}{2}\\4x^2-15x+8=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{3}{2}\\\left[{}\begin{matrix}x=\dfrac{15+\sqrt{97}}{8}\left(l\right)\\x=\dfrac{15-\sqrt{97}}{8}\left(tm\right)\end{matrix}\right.\end{matrix}\right.\)

Với \(2x+2y-5=0\Rightarrow2y=5-2x\)

\(\rightarrow\sqrt{3x+1}=2x-2\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\3x+1=4x^2-8x+4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\4x^2-11x+3=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le1\\\left[{}\begin{matrix}x=\dfrac{11+\sqrt{73}}{8}\left(tm\right)\\x=\dfrac{11-\sqrt{73}}{8}\left(l\right)\end{matrix}\right.\end{matrix}\right.\)

a) \(\left(x+\dfrac{1}{2}\right)+\left(x+\dfrac{1}{6}\right)+\left(x+\dfrac{1}{12}\right)+....+\left(x+\dfrac{1}{9900}\right)\)

\(\Leftrightarrow\left(x+x+x+...+x\right)+\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{9900}\right)=1\)

\(\Leftrightarrow50x+\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)=1\)

\(\Leftrightarrow50x+\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)=1\)

\(\Leftrightarrow50x+\left(1-\dfrac{1}{100}\right)=1\)

\(\Leftrightarrow50x+\dfrac{99}{100}=1\)

\(\Leftrightarrow50x=\dfrac{1}{100}\Rightarrow x=\dfrac{1}{5000}\)

b) \(A=\dfrac{3^2}{1.4}+\dfrac{3^2}{4.7}+\dfrac{3^2}{7.10}+...+\dfrac{3^2}{202.205}\)

\(A=\dfrac{3^2}{3}\cdot\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{202}-\dfrac{1}{205}\right)\)

\(A=\dfrac{9}{3}\cdot\left(1-\dfrac{1}{205}\right)\)

\(A=\dfrac{9}{3}\cdot\dfrac{204}{205}=\dfrac{615}{205}\)

a) \(\left(x+\dfrac{1}{2}\right)+\left(x+\dfrac{1}{6}\right)+\left(x+\dfrac{1}{12}\right)+....+\left(x+\dfrac{1}{9900}\right)=1\)

\(\Leftrightarrow\left(x+x+x+...+x\right)+\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{9900}\right)=1\)

\(\Leftrightarrow\left(x+x+x+...+x\right)+\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)=1\)

Có tất cả : (99 - 1) : 1 + 1 = 99 (số x)

\(\Rightarrow99x+\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)=1\)

\(\Rightarrow99x+\left(1-\dfrac{1}{100}\right)=1\)

\(\Rightarrow99x+\dfrac{99}{100}=1\Rightarrow99x=1-\dfrac{99}{100}\)

\(\Rightarrow99x=\dfrac{1}{100}\Rightarrow x=\dfrac{1}{100.99}=\dfrac{1}{9900}\)

b) \(A=\dfrac{3^2}{1.4}+\dfrac{3^2}{4.7}+\dfrac{3^2}{7.10}+....+\dfrac{3^2}{202.205}\)

\(A=\dfrac{3^2}{3}\cdot\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{202}-\dfrac{1}{205}\right)\)

\(A=\dfrac{9}{3}\cdot\left(1-\dfrac{1}{205}\right)\)

\(A=3\cdot\dfrac{204}{205}=\dfrac{615}{205}\)

thiếu cái = 1 ở cuối nữa nhé

Gọi số học sinh nam của lớp 7A là a ; số học sinh nữ là b . Ta có :

\(\dfrac{a}{3}=\dfrac{b}{5}\) và a +b = 40

Áp dụng tính chất của dãy tỉ số bằng nhau ; ta được:

\(\dfrac{a}{3}=\dfrac{b}{5}=\dfrac{a+b}{3+5}=\dfrac{40}{8}=5\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a}{3}=5\\\dfrac{b}{5}=5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=15\\b=25\end{matrix}\right.\)

Vậy có 15 học sinh nam ; 25 học sinh nữ