Ta có: \(\left(a-b\right)^2\ge0\Leftrightarrow a^2+b^2\ge2ab\Leftrightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2>100^2\)

\(\Leftrightarrow a^2+b^2>\frac{100^2}{2}=5000\)

e) \(AH\perp BC\)(giả thiết).

\(\Rightarrow\Delta HAB\)vuông tại H.

\(\Rightarrow S_{HAB}=\frac{AH.BH}{2}=4,8.\frac{30}{14}=\frac{144}{14}=\frac{72}{7}\left(cm^2\right)\)

Xét \(\Delta ABC\)có phân giác BD (giả thiết).

\(\Rightarrow\frac{AD}{CD}=\frac{AB}{BC}\)(tính chất).

\(\Rightarrow\frac{AD}{CD+AD}=\frac{AB}{BC+AB}\)(tính chất của tỉ lệ thức).

\(\Rightarrow\frac{AD}{AC}=\frac{AB}{BC+AB}\)

\(\Rightarrow\frac{AD}{8}=\frac{6}{10+6}=\frac{6}{16}=\frac{3}{8}\)(thay số).

\(\Rightarrow AD=\frac{3}{8}.8=3\left(cm\right)\)

Vì \(\Delta ABC\)vuông tại A (giả thiết).

\(\Rightarrow\widehat{CAB}=90^0\Rightarrow\widehat{DAB}=90^0\)

\(\Rightarrow\Delta ADB\)vuông tại A.

\(\Rightarrow S_{ADB}=\frac{AD.AB}{2}=\frac{3.6}{2}=9\left(cm^2\right)\)

Ta có: \(S_{ABC}=\frac{AB.AC}{2}\)(theo câu a))

\(\Rightarrow S_{ABC}=\frac{6.8}{2}=\frac{48}{2}=24\left(cm^2\right)\)

Lại có: \(S_{ABD}+S_{BCD}=S_{ABC}\)

\(\Rightarrow9+S_{BCD}=24\)(thay số).

\(\Rightarrow S_{BCD}=24-9=15\left(cm^2\right)\)

Vậy \(S_{HAB}=\frac{72}{7}cm^2;S_{BCD}=15cm^2\)

X+1/X  -1=3X-1/X+1  +1/X+1/

mình cần lời giải

244320035

.2356.050530568605203562.36.

ĐKXĐ : x ≠ 0 ; x ≠ -1

<=> \(\frac{2x+1}{x}=\frac{x\left(3x-1\right)}{x\left(x+1\right)}+\frac{1}{x\left(x+1\right)}\)

<=> \(\frac{2x+1}{x}=\frac{3x^2-x+1}{x\left(x+1\right)}\)

=> x( 2x + 1 )( x + 1 ) = x( 3x² - x + 1 )

<=> 2x³ + 3x² + x = 3x³ - x² + x 

<=> x³ - 4x² = 0

<=> x²( x - 4 ) = 0

<=> x = 0 (ktm) hoặc x = 4 (tm)

Vậy ...

Hoàng Trúc Nhi i don't need:))

\(\frac{x+1}{x}=\frac{x^2+1}{x^2}\)( ĐKXĐ : x ≠ 0 )

=> x( x² + 1 ) = x²( x + 1 )

<=> x³ + x = x³ + x²

<=> x³ + x² - x³ - x = 0

<=> x( x - 1 ) = 0

<=> x = 0(ktm) hoặc x = 1(tm)

Vậy phương trình có nghiệm x=1

giúp mik vs các bn ơi,mai mik cần gấp,bn nào nhanh nhất mik k cho 3 ngày liên tiếp nhá ! :3

a) ĐKXĐ: 

\(\hept{\begin{cases}x-3\ne0,9-x^2\ne0,x+3\ne0\\1-\frac{x+1}{x+3}\ne0\end{cases}}\Leftrightarrow x\ne\pm3\).

\(M=\left(\frac{3+x}{x-3}+\frac{18}{9-x^2}+\frac{x-3}{x+3}\right)\div\left(1-\frac{x+1}{x+3}\right)\)

\(M=\frac{\left(3+x\right)\left(x+3\right)-18+\left(x-3\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\div\frac{2}{x+3}\)

\(M=\frac{x^2+6x+9-18+x^2-6x+9}{\left(x-3\right)\left(x+3\right)}\times\frac{x+3}{2}\)

\(M=\frac{2x^2\left(x+3\right)}{2\left(x-3\right)\left(x+3\right)}\)

\(M=\frac{x^2}{x-3}\)

b) \(M=\frac{x^2}{x-3}=\frac{x^2-3x+3x-9+9}{x-3}=x+3+\frac{9}{x-3}\inℤ\Leftrightarrow\frac{9}{x-3}\inℤ\)

mà \(x\inℤ\)nên \(x-3\inƯ\left(9\right)=\left\{-9,-3,-1,1,3,9\right\}\Leftrightarrow x\in\left\{-6,0,2,4,6,12\right\}\). 

a) ĐKXĐ : x ≠ ±3

\(=\left[\frac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}-\frac{18}{\left(x-3\right)\left(x+3\right)}+\frac{\left(x-3\right)^2}{\left(x-3\right)\left(x+3\right)}\right]\div\left(\frac{x+3}{x+3}-\frac{x+1}{x+3}\right)\)

\(=\left[\frac{x^2+6x+9-18+x^2-6x+9}{\left(x-3\right)\left(x+3\right)}\right]\div\left(\frac{x+3-x-1}{x+3}\right)\)

\(=\frac{2x^2}{\left(x-3\right)\left(x+3\right)}\div\frac{2}{x+3}=\frac{2x^2}{\left(x-3\right)\left(x+3\right)}\times\frac{x+3}{2}=\frac{x^2}{x-3}\)

b) \(M=\frac{x^2}{x-3}=\frac{x^2-3x+3x-9+9}{x-3}=\frac{x\left(x-3\right)+3\left(x-3\right)+9}{x-3}=x+3+\frac{9}{x-3}\)

Vì x nguyên nên x + 3 nguyên

nên để M nguyên thì 9/x-3 nguyên

hay x - 3 ∈ Ư(9) [ bạn tự xét tiếp :)) ]

( x - 3 )( x + 4 ) - 2( 3x - 2 ) = ( x - 4 )²

<=> x² + x - 12 - 6x + 4 = x² - 8x + 16

<=> x² - 5x - x² + 8x = 24

<=> 3x = 24 <=> x = 8

Vậy pt có nghiệm x = 8

a) 2x + 4 = 7x - 2

<=> 2x - 7x = -2 - 4

<=> -5x = -6 

<=> x = 6/5

Vậy ...

b) x² - 4 - ( x + 5 )( 2 - x ) = 0

<=> ( x - 2 )( x + 2 ) + ( x + 5 )( x - 2 ) = 0

<=> ( x - 2 )( 2x + 7 ) = 0

<=> x = 2 hoặc x = -7/2

Vậy ...

c) \(\frac{x+2}{x-2}-\frac{x-2}{x+2}=\frac{4}{\left(x-2\right)\left(x+2\right)}\left(ĐKXĐ:x\ne\pm2\right)\)

<=> \(\frac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}-\frac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}=\frac{4}{\left(x-2\right)\left(x+2\right)}\)

=> x² + 4x + 4 - x² + 4x - 4 = 4

<=> 8x = 4 <=> x = 1/2(tm)

Vậy ...

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