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Nãy ghi nhầm =="
a)Hđ gđ là nghiệm pt
`x^2=2x+2m+1`
`<=>x^2-2x-2m-1=0`
Thay `m=1` vào pt ta có:
`x^2-2x-2-1=0`
`<=>x^2-2x-3=0`
`a-b+c=0`
`=>x_1=-1,x_2=3`
`=>y_1=1,y_2=9`
`=>(-1,1),(3,9)`
Vậy tọa độ gđ (d) và (P) là `(-1,1)` và `(3,9)`
b)
Hđ gđ là nghiệm pt
`x^2=2x+2m+1`
`<=>x^2-2x-2m-1=0`
PT có 2 nghiệm pb
`<=>Delta'>0`
`<=>1+2m+1>0`
`<=>2m> -2`
`<=>m> 01`
Áp dụng hệ thức vi-ét:`x_1+x_2=2,x_1.x_2=-2m-1`
Theo `(P):y=x^2=>y_1=x_1^2,y_2=x_2^2`
`=>x_1^2+x_2^2=14`
`<=>(x_1+x_2)^2-2x_1.x_2=14`
`<=>4-2(-2m-1)=14`
`<=>4+2(2m+1)=14`
`<=>2(2m+1)=10`
`<=>2m+1=5`
`<=>2m=4`
`<=>m=2(tm)`
Vậy `m=2` thì ....
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a: AB/AC=5/6
=>HB/HC=25/36
=>HB/25=HC/36=k
=>HB=25k; HC=36k
AH^2=HB*HC
=>25k*36k=30^2
=>900k^2=900
=>k=1
=>x=25cm; y=25cm
b: AB/AC=3/4
=>HB/HC=9/16
=>x/y=9/16
=>x/9=y/16=(x+y)/(9+16)=125/25=5
=>x=45cm; y=80cm
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(5\sqrt{\dfrac{9x-27}{25}}-7\sqrt{\dfrac{4x-12}{9}}-7\sqrt{x^2-9}+18\sqrt{\dfrac{9x^2-81}{81}}=0\)
\(\Leftrightarrow5\cdot\dfrac{3\sqrt{x-3}}{5}-7\cdot\dfrac{2\sqrt{x-3}}{7}-7\cdot\sqrt{x^2-9}+18\cdot\dfrac{\sqrt{x^2-9}}{3}=0\)
\(\Leftrightarrow\sqrt{x-3}-\sqrt{x^2-9}=0\)
\(\Leftrightarrow\sqrt{x-3}\left(1-\sqrt{x+3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+3=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(nhận\right)\\x=-2\left(loại\right)\end{matrix}\right.\)
ĐK: \(x\ge3\)
\(5\sqrt{\dfrac{9x-27}{25}}-7\sqrt{\dfrac{4x-12}{9}}-7\sqrt{x^2-9}+18\sqrt{\dfrac{9x^2-81}{81}}=0\)
\(\Leftrightarrow3\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}-7\sqrt{x^2-9}+6\sqrt{x^2-9}=0\)
\(\Leftrightarrow-5\sqrt{x-3}-3\sqrt{x^2-9}=0\)
Ta thấy \(-5\sqrt{x-3}-3\sqrt{x^2-9}\le0\forall x\ge3\) nên phương trình tương đương:
\(\left\{{}\begin{matrix}\sqrt{x-3}=0\\\sqrt{x^2-9}=0\end{matrix}\right.\)
\(\Leftrightarrow x=3\left(tm\right)\)
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\(\sqrt{5}+\sqrt[5]{20}-\sqrt[3]{45}\)
=\(2,23+1,82-3,55\)
=\(0,5\)
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\(a,=2\cdot5-4=6\\ b,=5\sqrt{3}-6\sqrt{3}+2\sqrt{3}=\sqrt{3}\\ c,=\sqrt{5}+3-\sqrt{5}=3\\ d,=\sqrt{\left(2+\sqrt{5}\right)^2}-\sqrt{5}=2+\sqrt{5}-\sqrt{5}=2\)
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7:
a: \(P=\left(1:\dfrac{x-x+1}{\sqrt{x}+\sqrt{x-1}}-\dfrac{x-1-2}{\sqrt{x-1}-\sqrt{2}}\right)\cdot\left(\dfrac{2\sqrt{x}-\sqrt{x}-\sqrt{2}}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}\right)\)\(=\left(\sqrt{x}+\sqrt{x-1}-\sqrt{x-1}-\sqrt{2}\right)\cdot\dfrac{\sqrt{x}-\sqrt{2}}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}\)
\(=-\dfrac{\left(\sqrt{x}-\sqrt{2}\right)}{\sqrt{x}}\)
b: Khi x=3-2căn 2 thì \(P=-\dfrac{\sqrt{2}-1-\sqrt{2}}{\sqrt{2}-1}=\dfrac{1}{\sqrt{2}-1}=\sqrt{2}+1\)
đề là j bn :?
\(\dfrac{1}{1+\sqrt{2}+\sqrt{7}}=\dfrac{\sqrt{2}+1-\sqrt{7}}{3+2\sqrt{2}-7}\)
\(=\dfrac{\sqrt{2}-\sqrt{7}+1}{-4+2\sqrt{2}}=\dfrac{\left(\sqrt{2}-\sqrt{7}+1\right)\left(2+\sqrt{2}\right)}{-2\left(2-\sqrt{2}\right)\left(2+\sqrt{2}\right)}\)
\(=\dfrac{2\sqrt{2}+2-2\sqrt{7}-\sqrt{14}+2+2\sqrt{2}}{-4}\)
\(=\dfrac{4\sqrt{2}+4-2\sqrt{7}-\sqrt{14}}{-4}=\dfrac{-4\sqrt{2}-4+2\sqrt{7}+\sqrt{14}}{4}\)