cac ban giup mk cau nay nhe x ngũ 2 -5x+4=0
Những câu hỏi liên quan
(x+1).(x+3).(x+5)<0
cac ban lam on tra loi cau hoi nay giup mk nhe
cac ban giai giup minh cau nay voi
g) x(x-2)-x2 = 5x-7
H) 3x(x-7)+2(x-7)=0
g/ x(x-2)-x2=5x-7
\(\Leftrightarrow x^2-2x-x^2=5x-7\\ \Leftrightarrow7x=7\Leftrightarrow x=1\)
h/\(3x\left(x-7\right)+2\left(x-7\right)=0\\ \Leftrightarrow3x^2-19x-14=0\\ \)
\(\Leftrightarrow\left(x-7\right)\left(3x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=7\\x=-\frac{2}{3}\end{matrix}\right.\)
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cac ban giup mk cau nay giai pt v(5+v(5-x))=x
Bài 1:
CMR: a2+4b2+4c2>= 4ab-4ac+8bc ( với mọi abc)
cac ban oi bai nay hoi kho nhung cac ban co giup minh nhe. cau mong dieu hanh phuc se den voi nguoi giup minh lam bai nay.
a2+4b2+4c2>= 4ab-4ac+8bc
a2+4b2+4c2 - 4ab +4ac-8bc
(a2 - 4ab+4b2)+4c2+(4ac-8bc>=0)
suy ra (a-2b2)+2.2c.(a-2b)+(2c)2
(a-2b+2c)2>=0
dau = xảy ra khi va chỉ khi a+2c=2b
a2+4b2+4c2>= 4ab-4ac+8bc(dpcm)
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ban giai day du cho minh di. minh lam de nop ma
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CM: (a^2+b^2)^2>=ab(a+b)^2
Cac ban giup minh cau nay nhe! Minh giai cau nay ra roi, nhung cau nay lai khong co dieu kien a;b > 0 nen minh khong chac. Ban nao co cach ma khong dung toi dieu kien thi giup minh nhe!
P/s: Neu co ai giai ra (a-b)^2.(a^2+ab+b^2) giong minh thi chua chac da dung vi ngoac ( a^2 + ab + b^2 ) chua chac da duong ( ab chua chac da duong ). Minh cung khong biet thay cua minh quen ghi dieu kien hay de bai no nhu the nay nua!
bá tay luon,cá khi bá nốt chan
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\(a^2+ab+b^2=a^2+\frac{2.a.1}{2}b+\frac{1}{4}b^2+\frac{3}{4}b^2=\left(a+\frac{1}{2}b\right)^2+\frac{3}{4}b^2\ge0\)
\(\left(a^2+b^2\right)^2\ge ab.\left(a+b\right)^2\)
\(\Leftrightarrow a^4+2a^2b^2+b^4\ge ab.\left(a^2+2ab+b^2\right)\)
\(\Leftrightarrow a^4+2a^2b^2+b^4-a^3b-2a^2b^2-ab^3\ge0\)
\(\Leftrightarrow a^4-a^3b+b^4-ab^3\ge0\)
\(\Leftrightarrow a^{\text{3}}.\left(a-b\right)-b^3.\left(a-b\right)\ge0\)
\(\Leftrightarrow\left(a-b\right).\left(a-b\right).\left(a^2-ab+b^2\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2.\left(a^2-ab+b^2\right)\ge0\text{ vì }\hept{\begin{cases}\left(a-b\right)^2\ge0\\\left(a^2-ab+b^2\right)\ge0\left(cmt\right)\end{cases}}\)
Vì BĐT cuối đúng nên BĐT đầu đúng (đpcm)
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tim x biet :2*x-12*x=0. cac ban oi giup minh giai cau nay voi
2.x - 12.x = 0
=> 2 - 12 . x = 0
=> - 10.x = 0
=> x = 0 : ( - 10 )
=> x = 0
????
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\(2\times x-12\times x=0\)
\(\left(2-12\right)\times x=0\)
\(-10\times x=0\)
\(x=0:\left(-10\right)\)
\(x=0\)
Vậy x = 0
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Tim x :
X - 45 = 23
Bn nao lp 2 thi moi dc tra loi cau hoi nay
Mk kb vs cac ban lm chinh xac cau hoi nay
Mk lp 3 day nhe ! 😊😊😊
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☹➞cac ban giup minh cau nay voi
☛ tinh tong
M = 1+\(\dfrac{1}{5}\)+\(\dfrac{3}{35}\)+...\(\dfrac{3}{9603}\)+\(\dfrac{3}{9999}\)
➜giup minh nhanh nhe minh dang can gap lam moi nguoi a
➞thank you cac ban ❕
\(M=1+\dfrac{1}{5}+\dfrac{3}{35}+...+\dfrac{3}{9999}\\ =\dfrac{3}{3}+\dfrac{3}{15}+\dfrac{3}{35}+...+\dfrac{3}{9999}\\ =\dfrac{3}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{99\cdot101}\right)\\ =\dfrac{3}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\\ =\dfrac{3}{2}\left(1-\dfrac{1}{101}\right)=\dfrac{3}{2}\cdot\dfrac{100}{101}=\dfrac{150}{101}\)
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Rut gon phan thuc: (x^(3)-x^(2)-x-2)/(x^(5)-3x^(4)+4x^(3)-5x^(2)+3x-2)
Minh that su da bo tay vs bai tap nay roi! Cac ban hay giup minh nhe! Minh xin cam on!
\(\frac{x^3-x^2-x-2}{x^5-3x^4+4x^3-5x^2+3x-2}\)
\(=\frac{x^3-2x^2+x^2-2x+x-2}{x^5-2x^4-x^4+2x^3+2x^3-4x^2-x^2+2x+x-2}\)
\(=\frac{\left(x^3-2x^2\right)+\left(x^2-2x\right)+\left(x-2\right)}{\left(x^5-2x^4\right)-\left(x^4-2x^3\right)+\left(2x^3-4x^2\right)-\left(x^2-2x\right)+\left(x-2\right)}\)
\(=\frac{x^2\left(x-2\right)+x\left(x-2\right)+\left(x-2\right)}{x^4\left(x-2\right)-x^3\left(x-2\right)+2x^2\left(x-2\right)-x\left(x-2\right)+\left(x-2\right)}\)
\(=\frac{\left(x-2\right)\left(x^2+x+1\right)}{\left(x-2\right)\left(x^4-x^3+2x^2-x+1\right)}=\frac{x^2+x+1}{x^4-x^3+2x^2-x+1}\)
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