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[ COMBO ] BỒI DƯỠNG TOÁN 8 NÂNG CAO VÀ PHÁT TRIỂN (VŨ HỮU BÌNH-NXBGD) & TUYỂN TẬP ĐỀ THI HSG TOÁN 8 (NGUYỄN VĂN TÚ-THCS THANH MỸ)

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Nơi bồi dưỡng kiến thức Toán - Lý - Hóa cho học sinh cấp 2+3 /<br />

Diễn Đàn Toán - Lý - Hóa 1000B Trần Hưng Đạo Tp.Quy Nhơn Tỉnh Bình Định<br />

Tuyển tập đề thi <strong>HSG</strong> Toán 8<br />

2x<br />

3 3<br />

Cho biÓu thøc A =<br />

2 + x +<br />

2x<br />

+ 1<br />

C©u 5 ( 1 ®iÓm)<br />

. T×m gi¸ trÞ nguyªn cña x ®Ó A nhËn gi¸ trÞ nguyªn<br />

3 3 3<br />

3<br />

3<br />

a, Chøng minh r»ng x + y + z = ( x + y) − 3xy.<br />

( x + y) + z<br />

1 1 1<br />

b, Cho + + = 0.<br />

x<br />

C©u 1<br />

a, x # 2 , x # -2 , x # 0<br />

b , A =<br />

=<br />

=<br />

y<br />

⎛<br />

⎜<br />

⎝ x<br />

2<br />

x −<br />

z<br />

yz<br />

x<br />

xz<br />

y<br />

xy<br />

z<br />

TÝnh A = + +<br />

2 2 2<br />

x 2 1 ⎞ 6<br />

+ + ⎟ :<br />

− 4 2 − x x + 2 ⎠ x + 2<br />

2( x + 2)<br />

+ x − 2 6<br />

:<br />

( x − 2)( x + 2) x + 2<br />

− 6<br />

c, §Ó A > 0 th× > 0<br />

C©u 2 . §KX§ :<br />

x + 2<br />

. =<br />

§¸p ¸n<br />

( x − 2)( x + 2) 6 2 − x<br />

1<br />

⇔ 2 − x > 0 ⇔ x < 2<br />

2 − x<br />

1<br />

x ≠ −1;<br />

x ≠ −<br />

2<br />

2<br />

2<br />

x − 4x<br />

+ 1 x − 5x<br />

+ 1 x<br />

PT ⇔ + 1+<br />

+ 1 = 0 ⇔<br />

x + 1 2x<br />

+ 1<br />

⇔<br />

1<br />

2<br />

− 3x<br />

+ 2 x<br />

+<br />

x + 1<br />

2<br />

− 3x<br />

+ 2<br />

= 0<br />

2x<br />

+ 1<br />

2 ⎛ 1 1 ⎞<br />

2<br />

( x − 3x<br />

+ 2) ⎜ + ⎟ = 0 ⇔ ( x − 3x<br />

+ 2)( 3x<br />

+ 2) = 0 ⇔ ( x −1)( x − 2)( 3x<br />

+ 2) = 0<br />

⎝ x + 1<br />

2x<br />

+ 1⎠<br />

⇔ x =1 ; x = 2 ; x = - 2/ 3<br />

C 3 gi¸ trÞ trªn ®Òu tháa m·n §KX§ .<br />

⎧ 2⎫<br />

VËy PT ®· cho cã tËp nghiÖm S = ⎨1<br />

;2;− ⎬<br />

⎩ 3 ⎭<br />

C©u 3:<br />

1, ∆ ADQ = ∆ ABR v× chóng lµ hai tam gi¸c<br />

vu«ng (®Ó ý gãc cã c¹nh vu«ng gãc) vµ DA=BD<br />

( c¹nh h×nh vu«ng). Suy ra AQ=AR, nªn ∆ AQR<br />

lµ tam gi¸c vu«ng c©n. Chøng minh tîng tù ta<br />

cã: ∆ ARP= ∆ ADS<br />

do ®ã AP = AS vµ ∆ APS lµ tam gi¸c c©n t¹i A.<br />

2, AM vµ AN lµ ®êng trung tuyÕn cña tam gi¸c<br />

vu«ng c©n AQR vµ APS nªn AN ⊥ SP vµ<br />

AM ⊥ RQ.<br />

MÆt kh¸c : ∠ PAN = ∠PAM<br />

= 45 0 nªn gãc<br />

DIỄN ĐÀN <strong>TOÁN</strong> - LÍ - HÓA 1000B TRẦN HƯNG ĐẠO TP.QUY NHƠN<br />

MAN vu«ng. VËy tø gi¸c AHMN cã ba gãc vu«ng, nªn nã lµ h×nh ch÷ nhËt.<br />

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9<br />

Trường <strong>THCS</strong> Thanh Mỹ<br />

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