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(ĐÂY CHỈ LÀ CÁCH CỦA MÌNH THÔI NHA)
d)
Gọi x là độ dài của MN.
Ta có: AH = AK + KH (gt)
=> KH = AH -AK
hay KH = 9,6-3,6 =6
Ta có: SABC = SAMN + SMNBC (gt)
hay \(\dfrac{AK.MN}{2}+\dfrac{KH\left(BC+MN\right)}{2}\) = \(\dfrac{AB.AC}{2}\)
hay \(\dfrac{3,6.x}{2}+\dfrac{6\left(x+20\right)}{2}=\dfrac{12.16}{2}=96\)
\(\Leftrightarrow\) 3,6x + 6x + 120 = 96.2 = 192
\(\Leftrightarrow\) 9.6x = 192 - 120= 72
\(\Leftrightarrow\) x = \(\dfrac{72}{9,6}=7,5\)
SMNCB= \(\dfrac{KH\left(MN.BC\right)}{2}=\dfrac{6\left(7,5+20\right)}{2}=82,5\) (cm2)
B A C H 20 12 16 k AK=6 AH=9,6 M N MN // BC
đề giống bọn mk này
bạn tính diện tích ABC xong trừ đi diện tích AMN là ra kết quả là 82,5
b)x3-2x2-4xy2+x
=x(x2-2x-4y2+1)
=x[(x2-2x+1)-4y2]
=x[(x-1)2-4y2]
=x(x-1-2y)(x-1+2y)
c) (x+2)(x+3)(x+4)(x+5)-8
=[(x+2)(x+5)][(x+3)(x+4)]-8
=(x2+5x+2x+10)(x2+4x+3x+12)-8
=(x2+7x+10)(x2+7x+12)-8
đặt x2+7x+10 =a ta có
a(a+2)-8
=a2+2a-8
=a2+4a-2a-8
=(a2+4a)-(2a+8)
=a(a+4)-2(a+4)
=(a+4)(a-2)
thay a=x2+7x+10 ta đc
(x2+7x+10+4)(x2+7x+10-2)
=(x2+7x+14)(x2+7x+8)
bài 2 x3-x2y+3x-3y
=(x3-x2y)+(3x-3y)
=x2(x-y)+3(x-y)
=(x-y)(x2+3)
a) x3 - 4x2 + 4x
= x(x2 - 4x + 4)
= x(x - 2)2
b) x2 - 3x + 2
= x2 - x - 2x + 2
= (x2 - x) + (2x - 2)
= x(x - 1) + 2(x - 1)
= (x + 2)(x - 1)
c) 8x3 + \(\dfrac{1}{27}\)
= \(\left(2x+\dfrac{1}{3}\right)\)\(\left(4x^2-\dfrac{2}{3}x+\dfrac{1}{9}\right)\)
d) 64x3 - \(\dfrac{1}{8}\)
= \(\left(4x+\dfrac{1}{2}\right)\left(16x^2-2x+\dfrac{1}{4}\right)\)
e) x2 - 4 + (x - 2)2
= (x + 2)(x - 2) - (x - 2)2
= (x - 2)[(x + 2) - (x - 2)]
= (x - 2)(x + 2 - x + 2)
= 4(x - 2)
f) x3 - 2x3 + x - xy2
= -x3 + x - xy2
= -x(x2 - 1 + y2)
g) x3 - 4x2 - 12x + 27
= (x3 + 27) - (4x2 + 12x)
= (x + 3)(x2 - 3x + 9) - 4x(x + 3)
= (x + 3)[(x2 - 3x + 9) - 4x]
= (x + 3)(x2 - 3x + 9 - 4x)
= (x + 3)(x2 - 7x + 9)
h) 2x - 2y - x2 + 2xy - y2
= (2x - 2y) - (x2 - 2xy + y2)
= 2(x - y) - (x - y)2
= (x - y)(2 - x + y)
i) 3x2 + 6x + 3 - 3y2
= 3(x2 + 2x + 1 - y2)
= 3[(x2 + 2x + 1) - y2]
= 3[(x + 1)2 - y2]
= 3( x + 1 - y)(x + 1 + y)
k) 25 - x2 - y2 + 2xy
= 25 - (x2 - 2xy + y2)
= 25 - (x - y)2
= (5 - x + y)(5 + x - y)
l) 3x - 3y - x2 + 2xy - y2
= (3x - 3y) - (x2 - 2xy + y2)
= 3(x - y) - (x - y)2
= (x - y)(3 - x + y)
m) x2 - y2 + 2x - 2y
= (x2 - y2) + (2x - 2y)
= (x - y)(x + y) + 2(x - y)
= (x - y)(x + y + 2)
n) x4 + 2x3 - 4x - 4
= (x4 - 4) + (2x3 - 4x)
= (x2 - 2)(x2 + 2) + 2x(x2 - 2)
= (x2 - 2)(x2 + 2 + 2x)
o) x2(1 - x2) - 4x - 4x2
= x2(1 - x)( 1 + x) - 4x(1 + x)
= x(1 + x)[x(1 - x) - 4x]
= x(x + 1)(x - x2 - 4)
p) x3 + y3 + z3 - 3xyz
= x3 + y3 + z3 - 3x2y + 3x2y - 3xy2 + 3xy2 - 3xyz
= [(x3 + 3x2y + 3xy2 + y3) + z3] - (3x2y + 3xy2 + 3xyz)
= [(x + y)3 + z3] - 3xy(x + y + z)
= (x + y + z)[(x + y)2 - (x + y)z + z2] - 3xy(x + y + z)
= (x + y + z)(x2 + 2xy + y2 - xz - yz + z2 - 3xy)
= (x + y + z)(x2 + y2 + z2 - xy - xz - yz)
q) (x - y)3 + (y - z)3 + (z - x)3
= [(x - y) + (y - z)][(x - y)2 - (x - y)(y - z) + (y - z)2] + (z - x)3
= (x - z)(x2 - 2xy + y2 - xy + xz - y2 + yz + y2 - 2yz + z2) - (x - z)3
= (x - z)(x2 + y2 + z2 - 3xy + xz - yz) - (x - z)3
= (x - z)[x2 + y2 + z2 - 3xy + xz - yz - (x - z)2]
= (x - z)(x2 + y2 - 3xy + xz - yz - x2 + 2xz - z2)
= (x - z)(y2 - 3xy + 3xz - yz)
= (x - z)[(y2 - yz) - (3xy - 3xz)]
= (x - z)[y(y - z) - 3x(y - z)
= (x - z)(y - 3x)(y - z)
Nhớ tik nha
a)\(2x^2-7xy+5y^2\)
\(=2x^2-2xy-5xy+5y^2\)
\(=2x\left(x-y\right)-5y\left(x-y\right)\)
\(=\left(x-y\right)\left(2x-5y\right)\)
b)\(x^3+3x^2y-4xy^2-12y^3\)
\(=\left(x^3+3x^2y\right)-\left(4xy^2+12y^3\right)\)
\(=x^2\left(x+3y\right)-4y^2\left(x+3y\right)\)
\(=\left(x+3y\right)\left(x^2-4y^2\right)\)
\(=\left(x+3y\right)\left(x-2y\right)\left(x+2y\right)\)
1) \(\frac{x-y}{z-y}=-10\Leftrightarrow x-y=10\left(y-z\right)\)
\(\Leftrightarrow x-y=10y-10z\)
\(\Leftrightarrow x=11y-10z\)
Thay x=11y-10z vào biểu thức \(\frac{x-z}{y-z}\), ta có:
\(\frac{11y-10z-z}{y-z}=\frac{11y-11z}{y-z}=\frac{11\left(y-z\right)}{y-z}=11\)
Chá quá, có ghi nhìn không rõ đề
2) \(2x^2=9x-4\)
\(\Leftrightarrow2x^2-9x+4=0\)
\(\Leftrightarrow2x^2-8x-x+4=0\)
\(\Leftrightarrow2x\left(x-4\right)-1\left(x-4\right)\)
\(\Leftrightarrow\left(2x-1\right)\left(x-4\right)=0\)
\(\Leftrightarrow2x-1=0\) hoặc x-4=0
1) 2x-1=0<=>x=1/2
2)x-4=0<=>x=4(Loại)
=> x=1/2
Bài 4:
a: ĐKXĐ: \(x\notin\left\{1;-1\right\}\)
b: \(M=\dfrac{x}{2x-2}+\dfrac{x^2+1}{2-2x^2}\)
\(=\dfrac{x}{2\left(x-1\right)}-\dfrac{x^2+1}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^2+x-x^2-1}{2\left(x-1\right)\left(x+1\right)}=\dfrac{1}{2\left(x+1\right)}\)
c: Để M=1/2 thì 2(x+1)=2
=>x+1=1
hay x=0
Hình bạn tự vẽ nhé!!!
Ta có: \(\widehat{ACB}=180^o-\widehat{ACD}=180^o-100^o=80^o\\ \)
Xét tam giác ADC ta có: \(\widehat{DAC}+\widehat{ACD}+\widehat{ADC}=180^o\)
\(\Leftrightarrow y^o+100^o+x^o=180^o\)
\(\Leftrightarrow x^o+y^o=180^o-100^o=80^o\left(1\right)\)
Xét tam giác ABC ta có:\(\widehat{BAC}+\widehat{ABD}+\widehat{ADB}=180^o\)
\(\Leftrightarrow2y^o+2x^o+x^o=180^o\)
\(\Leftrightarrow2y^o+3x^o=180^o\left(2\right)\)
Thế (1) vào (2) ta được: \(2.\left(80-x^o\right)+3x^o=180^o\)
\(\Leftrightarrow160^o-2x^o+3x^o=180^o\)
\(\Leftrightarrow160^o+x^o=180^o\)
\(\Leftrightarrow x^o=180^o-160^o=20^o\)
Khi đó giá trị của \(x=20\)
Chúc bạn học tốt