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Câu 8: Tìm x?
\(\left(x+5\right)^2-\left(x+2\right)\left(x-3\right)=-2\\ \\ < =>x^2+10x+25-x^2+3x-2x+6=-2\\ < =>x^2-x^2+10x+3x-2x=-2-25-6\\ < =>11x=-33\\ =>x=-\frac{33}{11}=-3\)
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 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 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 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:
\(\frac{x+1}{2002}+\frac{x+2}{2001}+\frac{x+3}{2000}=\frac{x+4}{1999}+\frac{x+5}{1998}+\frac{x+6}{1997}\)
\(\Rightarrow\left(1+\frac{x+1}{2002}\right)+\left(1+\frac{x+2}{2001}\right)+\left(1+\frac{x+3}{2000}\right)=\left(1+\frac{x+4}{1999}\right)+\left(1+\frac{x+5}{1998}\right)+\left(1+\frac{x+6}{1997}\right)\)
\(\Rightarrow\frac{x+2003}{2002}+\frac{x+2003}{2001}+\frac{x+2003}{2000}=\frac{x+2003}{1999}+\frac{x+2003}{1998}+\frac{x+2003}{1997}\)
\(\Rightarrow\frac{x+2003}{2002}+\frac{x+2003}{2001}+\frac{x+2003}{2000}-\frac{x+2003}{1999}-\frac{x+2003}{1998}-\frac{x+2003}{1997}=0\)
\(\Rightarrow\left(x+2003\right)\left(\frac{1}{2002}+\frac{1}{2001}+\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}-\frac{1}{1997}\right)=0\)
Mà \(\frac{1}{2002}+\frac{1}{2001}+\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}-\frac{1}{1997}\ne0\)
\(\Rightarrow x+2003=0\)
\(\Rightarrow x=-2003\)
Vậy x = -2003
Câu 6:
Giải:
Áp dụng định lí Py-ta-go vào \(\Delta ABC\left(\widehat{B}=90^o\right)\) có:
\(\Rightarrow AB^2+BC^2=AC^2\)
\(\Rightarrow6^2+BC^2=10^2\)
\(\Rightarrow BC^2=64\)
\(\Rightarrow BC=8\)
\(\Rightarrow S_{ABCD}=8.6=48\left(cm^2\right)\)
Vậy...
Câu 1: 4cm
Câu 2: 6cm
Câu 3: 90o
Câu 4: -108
Câu 5: 2
Câu 6: 14
Câu 7: 43
Câu 8: -1
Câu 9: -3
Câu 10: -26
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
sao bạn làm được như thế vậy ?