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câu 7 mk bấm nhầm đáp án là 120
qua B kẻ đường thẳng song song với AM cắt AC ở N.
vì AM là phân giác góc BAC nên có :
\(\dfrac{AC}{AB}=\dfrac{CM}{BM}=\dfrac{12}{6}=2\) suy ra \(\dfrac{CM}{BC}=\dfrac{CM}{CM+BM}=\dfrac{12}{12+6}=\dfrac{2}{3}\)
vì AM song song với BN nên có :
1,\(\dfrac{CA}{AN}=\dfrac{CM}{BM}=\dfrac{12}{AN}=2\) suy ra AN=6
2,\(\dfrac{AM}{BN}=\dfrac{CM}{BC}=\dfrac{2}{3}=\dfrac{4}{BN}\)suy ra BN=6
vì AB=6 nên tam giác ABN đều
suy ra \(\widehat{NAB}\)=\(60^0\)
mà \(\widehat{NAB}+\widehat{BAC}=\)\(180^0\)
nên \(\widehat{BAC}=\)\(120^0\)
Evaluate the expression at
x3 + 12x + 48x + 64
= (x + 4)2
= (- 4 + 4)2
= 02
= 0
Fill in the blank: ............
x3 - a = (x - 2)(x2 + 2x + 4)
x3 - a = x3 - 8
a = 8
Evaluate
Answer:
a + b = 8
(a + b)2 = 82
a2 + b2 + 2ab = 64
a2 + b2 + 2 . 10 = 64
a2 + b2 + 20 = 64
a2 + b2 = 64 - 20
a2 + b2 = 44
(a - b)2
= a2 - 2ab + b2
= 44 - 2 . 10
= 44 - 20
= 24
Given .
Evaluate A at .
Answer: A
A = (x - 5)(x2 + 5x + 25) - x2(x + 3) + 3x2
= x3 - 125 - x3 - 3x2 + 3x2
= - 125
Given .
Evaluate A at .
Answer: A
Answer:
Given
Answer:
Câu 7:
Ta có: \(\left(x-3\right)^2\ge0\)
\(\Rightarrow A=\left(x-3\right)^2=21\ge21\)
Dấu " = " khi \(\left(x-3\right)^2=0\Rightarrow x-3=0\Rightarrow x=3\)
Vậy \(MIN_A=21\) khi x = 3
Câu 10:
\(A=4x^2+4x+11\\ =\left[\left(2x\right)^2+2.2x.1+1\right]+10\\ =\left(2x+1\right)^2+10\ge10\left(\forall x\in Z\right)\)
Vậy: \(Min_A=10\) khi \(x=-\frac{1}{2}\)
10) \(9x^2+4y^2=20xy\)
\(\Leftrightarrow\left(3x-2y\right)^2=8xy\)
\(\Rightarrow\left(3x-2y\right)=\sqrt{8xy}\)
--- \(9x^2+4y^2=20xy\)
\(\Leftrightarrow\left(3x+2y\right)^2=32xy\)
\(\Rightarrow\left(3x+2y\right)=\sqrt{32xy}\)
\(A=\frac{3x-2y}{3x+2y}=\frac{\sqrt{8xy}}{\sqrt{32xy}}=\frac{1}{2}=0,5\)
5) \(x^3+8-\left(x+2\right)\left(x^2+3x+3\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2-2x+4\right)-\left(x+2\right)\left(x^2+3x+3\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(-5x+1\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}x+2=0\Leftrightarrow x=-2\\-5x+1=0\Leftrightarrow x=0,2\end{matrix}\right.\)
Tổng các nghiệm là: -2+0,2=-1,8
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:
? 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
khó oa bạn