Rút gon \(\frac{48c^2+27b^2+36a^2}{a^2+b^2+c^2}\)
Những câu hỏi liên quan
Rút gọn :
a) a3 + 3a2 + 3a + b3 + 3b2 + 3b + 2
b) 27b3 + 54ab2 + 36a2b - 19a3 - 27a2 - 9a - 1
phân tích đa thức thành nhân tử:
8a3-36a2b+54ab2-27b3-8
8a3 - 36a2b + 54ab2 - 27b3 - 8
= ( 8a3 - 36a2b + 54ab2 - 27b3 ) - 8
= ( 2a - 3b )3 - 23
= ( 2a - 3b - 2 )[ ( 2a - 3b )2 + 2( 2a - 3b ) + 4 ]
= ( 2a - 3b - 2 )( 4a2 - 12ab + 9b2 + 4a - 6b + 4 )
Rút gon biểu thức
R = \(\frac{3a^2-2ab-b^2}{2a^2+ab-b^2}:\frac{3a^2-4ab+b^2}{3a^2+2ab-b^2}\)
\(R=\frac{3a^2-2ab-b^2}{2a^2+ab-b^2}:\frac{3a^2-4ab+b^2}{3a^2+2ab-b^2}\)
\(R=\frac{3a^2-2ab-b^2}{2a^2+ab-b^2}.\frac{3a^2+2ab-b^2}{3a^2-4ab+b^2}\)
\(R=\frac{\left(3a+b\right)\left(a-b\right)}{\left(a+b\right)\left(2a-b\right)}.\frac{\left(a+b\right)\left(3a-b\right)}{\left(a-b\right)\left(3a-b\right)}\)
\(R=\frac{3a+b}{2a-b}\)
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cho biểu thức A = \(\left(\frac{x}{x^2-4}-\frac{2}{x-2}+\frac{1}{x+12}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)
a) rút gon: x \(\ne\)+_ 2
b) tìm x để : A<0
c) Tìm x để A nguyên
Hình như đề sai.Sửa đề luôn nha !
\(ĐKXĐ:x\ne\pm2\)
\(A=\left(\frac{x}{x^2-4}-\frac{2}{x-2}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)
\(=\left(\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x-2\right)\left(x+2\right)}\right):\frac{\left(x-2\right)\left(x+2\right)+10-x^2}{x+2}\)
\(=\frac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}:\frac{x^2-4+10-x^2}{x+2}\)
\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}\cdot\frac{x+2}{-6}=\frac{1}{x-2}\)
b
Để \(A< 0\Rightarrow\frac{1}{x-2}< 0\Rightarrow x-2< 0\Rightarrow x< 2\)
c
Để A nguyên thì \(\frac{1}{x-2}\) nguyên
\(\Rightarrow1⋮x-2\)
\(\Rightarrow x-2\in\left\{1;-1\right\}\Rightarrow x\in\left\{3;1\right\}\)
Rút gon \(\left(\frac{2a}{a^2-4}+\frac{1}{2a}-\frac{2}{a+2}\right).\left(1+\frac{a^2+4}{4-a^2}\right)\)
Rút gon biểu thức
A =\(\left[\frac{1}{a^2}+\frac{1}{b^2}+\frac{2}{a+b}.\left(\frac{1}{a}+\frac{1}{b}\right)\right].\frac{ab}{\left(a+b\right)^2}\)
B = \(\left[\frac{1}{\left(2x-y\right)^2}+\frac{2}{4x^2-y^2}+\frac{1}{\left(2x+y\right)^2}\right].\frac{4x^2+4xy+y^2}{16x}\)
\(A=\left[\frac{1}{a^2}+\frac{1}{b^2}+\frac{2}{a+b}\left(\frac{1}{a}+\frac{1}{b}\right)\right].\frac{ab}{\left(a+b\right)^2}\)
\(=\left(\frac{1}{a}+\frac{1}{b}\right)^2.\frac{ab}{\left(a+b\right)^2}\)
\(=\frac{1}{ab}\)
\(B=\left[\frac{1}{\left(2x-y\right)^2}+\frac{2}{4x^2-y^2}+\frac{1}{\left(2x+y\right)^2}\right].\frac{4x^2+14xy+y^2}{16x}\)
\(=\frac{\left(2x+y\right)^2+2\left(2x+y\right)\left(2x-y\right)+\left(2x-y\right)^2}{\left(2x+y\right)^2.\left(2x-y\right)^2}.\frac{\left(2x+y\right)^2}{16x}\)
\(=\frac{\left(2x+y+2x-y\right)^2}{\left(2x+y\right)^2.\left(2x-y\right)^2}.\frac{\left(2x+y\right)^2}{16x}\)
\(=\frac{x}{\left(2x-y\right)^2}\)
\(A=\left[\frac{1}{a^2}+\frac{1}{b^2}+\frac{2}{a+b}.\left(\frac{1}{a}+\frac{1}{b}\right)\right].\frac{ab}{\left(a+b\right)^2}\)
ĐK: a, b khác 0, a khác -b
\(A=\left[\frac{1}{a^2}+\frac{1}{b^2}+\frac{2}{a+b}.\left(\frac{a+b}{ab}\right)\right].\frac{ab}{\left(a+b\right)^2}\)
\(A=\left[\frac{1}{a^2}+\frac{1}{b^2}+\frac{2}{ab}\right].\frac{ab}{\left(a+b\right)^2}=\left(\frac{1}{a}+\frac{1}{b}\right)^2.\frac{ab}{\left(a+b\right)^2}\)
\(A=\frac{\left(a+b\right)^2}{ab}.\frac{ab}{\left(a+b\right)^2}=1\)
\(B=\left[\frac{1}{\left(2x-y\right)^2}+\frac{2}{\left(4x^2-y^2\right)}+\frac{1}{\left(2x+y\right)^2}\right].\frac{4x^2+4xy+y^2}{16xy}\)
ĐK: xy khác 0, y \(\ne\pm\)2x
\(B=\left[\frac{1}{\left(2x-y\right)^2}+\frac{2}{\left(2x-y\right).\left(2x+y\right)}+\frac{1}{\left(2x+y\right)^2}\right].\frac{\left(2x+y\right)^2}{16xy}\)
\(B=\left[\frac{1}{\left(2x-y\right)}+\frac{1}{\left(2x+y\right)}\right]^2.\frac{\left(2x+y\right)^2}{16xy}\)
\(B=\left(\frac{2x+y+2x-y}{\left(2x-y\right).\left(2x+y\right)}\right)^2.\frac{\left(2x+y\right)^2}{16xy}\)
\(B=\frac{16x^2}{\left(2x-y\right)^2.\left(2x+y\right)^2}.\frac{\left(2x+y\right)^2}{16xy}\)
\(B=\frac{x}{\left(2x-y\right)^2.y}\)
Mình nhầm đề nhé. Làm lại như sau
\(A=\left[\frac{1}{a^2}+\frac{1}{b^2}+\frac{2}{a+b}.\left(\frac{1}{a}+\frac{1}{b}\right)\right].\frac{ab}{\left(a+b\right)^2}\)
\(A=\left[\frac{1}{a^2}+\frac{1}{b^2}+\frac{2}{a+b}.\left(\frac{a+b}{ab}\right)\right].\frac{ab}{\left(a+b\right)^2}\)
\(A=\left[\frac{1}{a^2}+\frac{1}{b^2}+\frac{2}{ab}\right].\frac{ab}{\left(a+b\right)^2}\)
\(A=\left[\frac{1}{a}+\frac{1}{b}\right]^2.\frac{ab}{\left(a+b\right)^2}\)=\(\frac{\left(a+b\right)^2}{\left(ab\right)^2}.\frac{ab}{\left(a+b\right)^2}=\frac{1}{ab}\)
\(B=\left[\frac{1}{\left(2x-y\right)^2}+\frac{2}{\left(2x+y\right).\left(2x-y\right)}+\frac{1}{\left(2x+y\right)^2}\right].\frac{4x^2+4xy+y^2}{16x}\)
\(B=\left[\frac{1}{\left(2x-y\right)}+\frac{1}{\left(2x+y\right)}\right]^2.\frac{\left(2x+y\right)^2}{16x}\)
\(B=\left[\frac{2x+y-2x-y}{\left(2x-y\right).\left(2x+y\right)}\right]^2.\frac{\left(2x+y\right)^2}{16x}\)
\(B=\frac{\left(4x\right)^2}{\left(2x-y\right)^2.\left(2x+y\right)^2}.\frac{\left(2x+y\right)^2}{16x}\)
\(B=\frac{16.x^2}{\left(2x-y\right)^2.\left(2x+y\right)^2}.\frac{\left(2x+y\right)^2}{16x}=\frac{x}{\left(2x-y\right)^2}\)
cho P=\(\left(\frac{2+x}{2-x}+\frac{4x^2}{x^2-4}-\frac{2-x}{2+x}\right):\frac{x^2-3x}{2x^2-x^3}\)
a) tìm điều kiện của x để giá trị S xác định
b) rút gon P
c)tính giá trị của P với |x-5|=2
Đk : \(\hept{\begin{cases}x-3\ne0\\x-2\ne0\\x+2\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne-2\\x\ne2\end{cases}}}\)\(P=\frac{\left(2+x\right)^2+4x^2-\left(2-x\right)^2}{\left(x-2\right)\left(x+2\right)}.\frac{x^2\left(2-x\right)}{x\left(x-3\right)}\)\(\Rightarrow P=\frac{8x+4x^2}{\left(x-2\right)\left(x+2\right)}.\frac{x\left(2-x\right)}{x-3}\)\(\Rightarrow p=\frac{4x\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}.\frac{x\left(x-2\right)}{3-x}=\frac{4x^2}{3-x}\)\(|x-5|=2\)nếu \(x\ge5\)=> x-5=2 =>x=7 (TM) => \(P=\frac{4.7^2}{-7+3}=-49\)Nếu \(x< 5\)=> x-5 = -2 => x = 3 Loại
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Rút gon biểu thức \(A=\frac{x^2+2x}{x^2-4x+4}:\left(\frac{x+2}{x}-\frac{1}{2-x}+\frac{6-x^2}{x^2-2x}\right)\)
ChoQleft(frac{sqrt{x}-2}{x-1}-frac{sqrt{x}+2}{x+2sqrt{x}+1}right)left(frac{1-x}{sqrt{2}}right)^2a, rút gọnb, chứng minh nếu 0x1 thì Q0c, tìm GTLN của Q ChoAfrac{1}{2left(1+sqrt{x}+2right)}+frac{1}{2left(1-sqrt{x}+2right)}a, tìm x để a có nghĩab, rút gon Ac, tìm X nguyên để A nguyên ChoAleft(frac{sqrt{a}}{sqrt{a-1}}-frac{1}{a-sqrt{a}}right):left(frac{1}{sqrt{a}-1}-frac{2}{a-1}right)a, Rút gọn Ab, tính A Khi a3+2sqrt{2}
Đọc tiếp
\(ChoQ=\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right)\left(\frac{1-x}{\sqrt{2}}\right)^2\)
a, rút gọn
b, chứng minh nếu 0<x<1 thì Q>0
c, tìm GTLN của Q
\(ChoA=\frac{1}{2\left(1+\sqrt{x}+2\right)}+\frac{1}{2\left(1-\sqrt{x}+2\right)}\)
a, tìm x để a có nghĩa
b, rút gon A
c, tìm X nguyên để A nguyên
\(ChoA=\left(\frac{\sqrt{a}}{\sqrt{a-1}}-\frac{1}{a-\sqrt{a}}\right):\left(\frac{1}{\sqrt{a}-1}-\frac{2}{a-1}\right)\)
a, Rút gọn A
b, tính A Khi a=3+\(2\sqrt{2}\)