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Câu 1:
\(\frac{x+3}{7}-\frac{5-x}{6}=\frac{37}{21}\)
\(\Rightarrow\frac{6\left(x+3\right)}{42}-\frac{7\left(5-x\right)}{42}=\frac{37}{21}\)
\(\Rightarrow\frac{6x+18}{42}-\frac{35-7x}{42}=\frac{37}{21}\)
\(\Rightarrow\frac{6x+18-35+7x}{42}=\frac{37}{21}\)
\(\Rightarrow\frac{13x-17}{42}=\frac{37}{21}\)
\(\Rightarrow21\left(13x-17\right)=42.37\)
\(\Rightarrow13x-17=2.37\)
\(\Rightarrow13x-17=74\)
\(\Rightarrow13x=91\)
\(\Rightarrow x=7\)
Vậy x = 7
Câu 4: Ta có: \(S_{ABC}=45cm^2\Rightarrow a.h=45.2=90cm\)
Mà \(BC=a;AH=10cm\)
\(\Rightarrow BC=90:AH=90:10=9cm\)
Vậy \(BC=9cm\)
Câu 10: (Txđ : a,b,c khác 0) a,b,c là số dương nên Cô-si được:
B=\(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{a}{c}+\dfrac{b}{a}+\dfrac{c}{b}+\dfrac{c}{a}\ge2\sqrt{\dfrac{a}{b}.\dfrac{b}{a}}+2\sqrt{\dfrac{b}{c}.\dfrac{c}{b}}+2\sqrt{\dfrac{a}{c}.\dfrac{c}{a}}=6\)
Dấu "=" xảy ra <=> a = b = c
Vậy min B = 6 khi a = b = c
Câu 1:
? 10cm H B A C
ta có: \(S_{ABC}=\dfrac{1}{2}.AH.BC\)
hay \(45=\dfrac{1}{2}.10.BC\)
\(\Rightarrow BC=\dfrac{45}{5}=9\)
Vậy BC = 9(cm)
Câu 1:
Độ dài BC bằng:
\(S_{ABC}=\frac{AH.BC}{2}\\ =>BC=\frac{S_{ABC}.2}{AH}=\frac{45.2}{10}=9\left(cm\right)\)
Câu 1:
Cạnh BC bằng:
\(S_{ABC}=\frac{AH.BC}{2}\\ =>BC=\frac{S_{ABC}.2}{AH}=\frac{45.2}{10}=9\left(cm\right)\)
Câu 6:
A B C D
Giải:
Xét \(\Delta ABC\left(\widehat{B}=90^o\right)\) có:
\(AB^2+BC^2=AC^2\)
\(\Rightarrow\sqrt{2^2}+\sqrt{2^2}=AC^2\)
\(\Rightarrow AC^2=4\)
\(\Rightarrow AC=2\)
Vậy đường chéo là 2 cm
Câu 1:
\(\dfrac{x+1}{2002}+\dfrac{x+2}{2001}+\dfrac{x+3}{2000}=\dfrac{x+4}{1999}+\dfrac{x+5}{1998}+\dfrac{x+6}{1997}\)
\(\Leftrightarrow\left(\dfrac{x+1}{2002}+1\right)+\left(\dfrac{x+2}{2001}+1\right)+\left(\dfrac{x+3}{2000}+1\right)=\left(\dfrac{x+4}{1999}+1\right)+\left(\dfrac{x+5}{1998}+1\right)+\left(\dfrac{x+6}{1997}+1\right)\) \(\Leftrightarrow\dfrac{x+2003}{2002}+\dfrac{x+2003}{2001}+\dfrac{x+2003}{2000}=\dfrac{x+2003}{1999}+\dfrac{x+2003}{1998}+\dfrac{x+2003}{1997}\) \(\Leftrightarrow\left(x+2003\right)\left(\dfrac{1}{2002}+\dfrac{1}{2001}+\dfrac{1}{2000}-\dfrac{1}{1999}-\dfrac{1}{1998}-\dfrac{1}{1997}\right)=0\)
\(\Leftrightarrow x+2003=0\)
\(\Leftrightarrow x=-2003\)
Vậy S={-2003}
Câu 9:
\(\left(x+5\right)^2-\left(x+2\right)\left(x-3\right)=-2\)
\(\Leftrightarrow x^2+10x+25-x^2-3x+2x+6+2=0\)
\(\Leftrightarrow9x+33=0\)
\(\Leftrightarrow9x=-33\)
\(\Leftrightarrow x=\dfrac{-11}{3}\)
Vậy S=\(\left\{\dfrac{-11}{3}\right\}\)
Câu 2:\(\dfrac{x+1}{2002}+\dfrac{x+2}{2001}+\dfrac{x+3}{2000}=\dfrac{x+4}{1999}+\dfrac{x+5}{1998}+\dfrac{x+6}{1997}\)
\(\Leftrightarrow\dfrac{x+1}{2002}+\dfrac{x+2}{2001}+\dfrac{x+3}{2000}-\dfrac{x+4}{1999}-\dfrac{x+5}{1998}-\dfrac{x+6}{1997}=0\)
\(\Leftrightarrow\dfrac{x+1}{2002}+1+\dfrac{x+2}{2001}+1+\dfrac{x+3}{2000}+1-\dfrac{x+4}{1999}+1-\dfrac{x+5}{1998}+1-\dfrac{x+6}{1997}+1\)
\(\Leftrightarrow\dfrac{x+2003}{2002}+\dfrac{x+2003}{2001}+\dfrac{x+2003}{2000}-\dfrac{x+2003}{1999}-\dfrac{x+2003}{1998}-\dfrac{x+2003}{1997}=0\)
\(\Leftrightarrow\left(x+2003\right)\left(\dfrac{1}{2002}+\dfrac{1}{2001}+\dfrac{1}{2000}-\dfrac{1}{1999}-\dfrac{1}{1998}-\dfrac{1}{1997}\ne0\right)=0\)
\(\Leftrightarrow x+2003=0\)
\(\Leftrightarrow x=-2003\)
Vậy PT có nghiệm là \(x=-2003\)
Bài 3: Tìm x?
\(\frac{x+3}{7}-\frac{5-x}{6}=\frac{37}{21}\\ < =>6\left(x+3\right)-7\left(5-x\right)=74\\ < =>6x+18-35+7x=74\\ < =>6x+7x=74-18+35\\ < =>13x=91\\ =>x=\frac{91}{13}=7\)
câu 4
\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+...+\dfrac{1}{\left(x+99\right)\left(x+100\right)}=\dfrac{k}{x\left(x+100\right)}\) =>\(\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-...+\dfrac{1}{x+99}-\dfrac{1}{x+100}=\dfrac{k}{x\left(x+100\right)}\) =>\(\dfrac{1}{x}-\dfrac{1}{x+100}=\dfrac{k}{x\left(x+100\right)}\)
=>\(\dfrac{x+100-x}{x\left(x+100\right)}=\dfrac{k}{x\left(x+100\right)}\)
=>x+100-x=k
=>k=100
\(\Delta AMC\) và \(\Delta ABC\) có chung chiều cao hạ từ C và đáy AM=\(\dfrac{2}{3}AB\) nên\(S_{AMC}=\dfrac{2}{3}S_{ABC}=\dfrac{2}{3}.54=36\left(cm^2\right)\)
\(\Delta AMC\) và \(\Delta AMN\) có chung chiều cao hạ từ M và đáy \(AN=\dfrac{1}{3}AC=>S_{AMN}=\dfrac{1}{3}S_{AMC}=\dfrac{1}{3}.36=12\left(cm^2\right)\) Vậy diện tích tam giác AMN=12(cm2) A B C M N
Bài 8:
\(\left|x-3\right|=y\Rightarrow A=y\left(2-y\right)\Leftrightarrow2y-y^2\)
\(A=1-\left(y^2-2y+1\right)=1-\left(y-1\right)^2\le1\)
GTNN(A)=1 khi y=1\(\Rightarrow\left[\begin{matrix}x=2\\x=4\end{matrix}\right.\)
Câu 8: Tìm x?
\(\left(x+5\right)^2-\left(x+2\right)\left(x-3\right)=-2\\ \\ < =>x^2+10x+25-x^2+3x-2x+6=-2\\ < =>x^2-x^2+10x+3x-2x=-2-25-6\\ < =>11x=-33\\ =>x=-\frac{33}{11}=-3\)