tìm x \(\sqrt{2x-1}\)-\(\sqrt{8x-4}\)+\(\sqrt{50x-25}\)=24
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\(\sqrt{x-1}\) - \(\sqrt{9x-9}\) + \(\sqrt{16x-16}\) = 4 (đk \(x\ge\)1)
\(\sqrt{x-1}-\) \(\sqrt{9\left(x-1\right)}\) + \(\sqrt{16\left(x-1\right)}\) = 4
\(\sqrt{x-1}\) - 3\(\sqrt{x-1}\) + \(4\sqrt{x-1}\) = 4
\(\sqrt{x-1}\)( 1 - 3 + 4 ) = 4
\(\sqrt{x-1}\) . 2 = 4
\(\sqrt{x-1}\) = 4 : 2
\(\sqrt{x-1}\) = 2
\(x-1\) =4
\(x=4+1\)
\(x=5\) (thỏa mãn)
Vậy \(x\) = 5
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\(\sqrt{3-x}\) - \(\sqrt{12-4x}\) + \(\sqrt{27-9x}\) = 20 đk \(3-x\) ≥ 0 ⇒ \(x\le3\)
\(\sqrt{3-x}\) - \(\sqrt{4.\left(3-x\right)}\) + \(\sqrt{9.\left(3-x\right)}\) = 20
\(\sqrt{3-x}\) - 2\(\sqrt{3-x}\) + 3\(\sqrt{3-x}\) = 20
\(\sqrt{3-x}\).( 1 - 2 + 3) = 20
2\(\sqrt{3-x}\) = 20
\(\sqrt{3-x}\) = 20: 2
\(\sqrt{3-x}\) = 10
3 - \(x\) = 100
\(x\) = 3 - 100
\(x\) = -97 (thỏa mãn)
Vậy \(x\) = -97
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\(\sqrt{x+3}\) + \(\sqrt{9x+27}\) - \(\sqrt{4x-12}\) = 10 đk \(x+3\) ≥ 0 ⇒ \(x\) ≥ -3
\(\sqrt{x+3}\) + \(\sqrt{9\left(x+3\right)}\) - \(\sqrt{4\left(x+3\right)}\) = 10
\(\sqrt{x+3}\) + 3\(\sqrt{x+3}\) - 2\(\sqrt{x+3}\) = 10
(1 + 3 - 2)\(\sqrt{x+3}\) = 10
2\(\sqrt{x+3}\) = 10
\(\sqrt{x+3}\) = 10: 2
\(\sqrt{x+3}\) = 5
\(x+3\) = 10
\(x\) = 10 - 3
\(x\) = 7 ( thỏa mãn)
Vậy \(x\) = 7
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\(\sqrt{x+2}\) + \(\sqrt{16x+32}\) - \(\sqrt{4x+8}\) = 16 (đk \(x\ge\) -2)
\(\sqrt{x+2}\) + \(\sqrt{16\left(x+2\right)}\) - \(\sqrt{4\left(x+2\right)}\) = 16
\(\sqrt{x+2}\) + 4\(\sqrt{x+2}\) - 2\(\sqrt{x+2}\) = 16
( 1 + 4 - 2)\(\sqrt{x+2}\) = 16
3\(\sqrt{x+2}\) = 16
\(\sqrt{x+2}\) = \(\dfrac{16}{3}\)
\(x+2\) = \(\dfrac{256}{9}\)
\(x\) = \(\dfrac{256}{9}\) - 2
\(x\) = \(\dfrac{238}{9}\) (thỏa mãn)
Vậy \(x=\dfrac{238}{9}\)
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Ta thấy \(A-4=\dfrac{x+2\sqrt{x}+1}{\sqrt{x}}-4\)
\(=\dfrac{x+2\sqrt{x}+1-4\sqrt{x}}{\sqrt{x}}\)
\(=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}}\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}\)
Do \(\left(\sqrt{x}-1\right)^2\ge0\) và \(\sqrt{x}>0\) nên \(\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}\ge0\). ĐTXR \(\Leftrightarrow x=1\).
Như vậy \(A-4\ge0\) \(\Leftrightarrow A\ge4\)
(không phải là \(A>4\) như trong đề đâu nhé, dấu "=" vẫn có thể xảy ra nếu \(x=1\))
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1 was asked to type the report by the boss..
Bị động thì quá khứ đơn: S + was/were + PII + (by O)......
2 Lily is suspected to have stolen all the books in the library.
Câu bị động khách quan
S+report/say/said/tell/told/........(V1)+That+S2+V2
=It + "be"+reported/said/told/.......(V1)+that+S2+V2
=S2 +"be"+Reported/said/told/......(V1)+to V(khi V1 và V2 cùng thì) /to have PII(khi V1 và V2 khác thì)
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1 break -> to break
Đảo ngữ câu điều kiện loại 2: Were + S + Adj/N/to V, S + would/could + V(infinitive)
2 have been -> be
Câu điều kiện hỗn hợp loại 3 và 2: If + S + V(quá khứ hoàn thành), S + would/could + V (infinitive)
3 will -> would
Câu điều kiện loại 2 diễn tả điều không thể xảy ra ở hiện tại hoặc tương lai: If + S+ V(quá khứ đơn), S + would/could + V (infinitive)
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1. 'You look sad – 'I _________ about our breakup this morning.'
A. just think В. had just thought
С. was just thinking D. just thought
2. It was the first time I______________ a live Black Pink concert.
A. was ever seeing C. had ever seen
B. had ever been seeing D. have ever seen
3. I________________ the whole book by the time Linda calls.
A. was reading C. had read
B. had been reading D. will have read
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\(\sqrt{2x-1}\) - \(\sqrt{8x-4}\) + \(\sqrt{50x-25}\) = 24 đk \(x\ge\dfrac{1}{2}\)
\(\sqrt{2x-1}\) - \(\sqrt{4.\left(2x-1\right)}\) + \(\sqrt{25.\left(2x-1\right)}\) = 24
\(\sqrt{2x-1}\) - 2\(\sqrt{2x-1}\) + 5\(\sqrt{2x-1}\) = 24
\(\sqrt{2x-1}\) (1 - 2 + 5) = 24
4\(\sqrt{2x-1}\) = 24
\(\sqrt{2x-1}\) = 24: 4
\(\sqrt{2x-1}\) = 6
\(2x-1=36\)
2\(x\) = 37
\(x=\dfrac{37}{2}\) (thỏa mãn)
Vậy \(x=\dfrac{37}{2}\)