Question

Ai dậy r giúp vs :33 1 câu cx đc nhé :v toàn giải pt hết nhé

1) \(4x^2+\left(8x-4\right)\sqrt{x}-1=3x+2\sqrt{2x^2+5x-3}.\)

2) \(\left(5x+8\right)\sqrt{2x-1}+7x\sqrt{x+3}=9x+18-\left(x+26\right)\sqrt{x-1}\)

3) \(\sqrt[3]{3x+1}+\sqrt[3]{5-x}+\sqrt[3]{2x-9}-\sqrt[3]{4x-3}=0\)

4) \(\left(x+17\right)\sqrt{4-x}+\left(20-x\right)\sqrt{x+1}-9\sqrt{4-x}.\sqrt{x+1}=36\)

Answer 1

Câu 1 là \(\left(8x-4\right)\sqrt{x}-1\) hay là \(\left(8x-4\right)\sqrt{x-1}\)?

Answer 2

Câu 1:ĐK \(x\ge\frac{1}{2}\)

\(4x^2+\left(8x-4\right)\sqrt{x}-1=3x+2\sqrt{2x^2+5x-3}\)

<=> \(\left(4x^2-3x-1\right)+4\left(2x-1\right)\sqrt{x}-2\sqrt{\left(2x-1\right)\left(x+3\right)}\)

<=> \(\left(x-1\right)\left(4x+1\right)+2\sqrt{2x-1}\left(2\sqrt{x\left(2x-1\right)}-\sqrt{x+3}\right)=0\)

<=> \(\left(x-1\right)\left(4x+1\right)+2\sqrt{2x-1}.\frac{8x^2-4x-x-3}{2\sqrt{x\left(2x-1\right)}+\sqrt{x+3}}=0\)

<=>\(\left(x-1\right)\left(4x+1\right)+2\sqrt{2x-1}.\frac{\left(x-1\right)\left(8x+3\right)}{2\sqrt{x\left(2x-1\right)}+\sqrt{x+3}}=0\)

<=> \(\left(x-1\right)\left(4x+1+2\sqrt{2x-1}.\frac{8x+3}{2\sqrt{x\left(2x-1\right)}+\sqrt{x+3}}\right)=0\)

Với \(x\ge\frac{1}{2}\)thì \(4x+1+2\sqrt{2x-1}.\frac{8x-3}{2\sqrt{x\left(2x-1\right)}+\sqrt{x+3}}>0\)

=> \(x=1\)(TM ĐKXĐ)

Vậy x=1

Answer 3

câu 2 ĐK \(x\ge1\)

\(\left(5x+8\right)\sqrt{2x-1}+7x\sqrt{x+3}=9x+18-\left(x+26\right)\sqrt{x-1}=0\)

<=> \(\left(5x+8\right)\left(\sqrt{2x-1}-1\right)+7x\left(\sqrt{x+3}-2\right)+\left(x+26\right)\sqrt{x-1}+10\left(x-1\right)=0\)

<=>\(\left(5x+8\right).\frac{2x-2}{\sqrt{2x-1}+1}+7x.\frac{x+3-4}{\sqrt{x+3}+2}+\left(x+26\right)\sqrt{x-1}+10\left(x-1\right)=0\)

<=> \(\sqrt{x-1}\left(\frac{2\left(5x+8\right)\sqrt{x-1}}{\sqrt{2x-1}+1}+\frac{7x\sqrt{x-1}}{\sqrt{x+3}+2}+\left(x+26\right)+10\sqrt{x-1}\right)=0\)

Với \(x\ge1\)thì cái trong ngoặc >0

=> \(x=1\)

Vậy x=1

Answer 4

Câu 3 . Bạn chuyển vế rồi lập phương là ra ngay.

Answer 5

3) \(\sqrt[3]{3x+1}+\sqrt[3]{5-x}+\sqrt[3]{2x-9}-\sqrt[3]{4x-3}=0\)

<=> \(\sqrt[3]{3x-1}+\sqrt[3]{5-x}=-\sqrt[3]{2x-9}+\sqrt[3]{4x-3}\)

<=> \(3x+1+5-x+3\sqrt[3]{\left(3x+1\right)^2}.\sqrt[3]{5-x}+3\sqrt[3]{\left(3x+1\right).\left(5-x\right)}\)\(=4x-3-2x+9-3\sqrt[3]{\left(4x-3\right)^2.\left(2x-9\right)}+3\sqrt[3]{\left(4x-3\right).\left(2x-9\right)^2}\)

<=> \(\sqrt[3]{\left(3x+1\right).\left(5-x\right)}.\left(\sqrt[3]{3x+1}+\sqrt[3]{5-x}\right)\)  \(=\sqrt[3]{\left(4x-3\right).\left(2x-9\right)}.\left(\sqrt[3]{2x-9}-\sqrt[3]{4x-3}\right)\)

<=> \(\sqrt[3]{\left(3x+1\right).\left(5-x\right)}=-\sqrt[3]{\left(4x-3\right).\left(2x-9\right)}\)

<=> (3x+`1).(5-x ) = - ( 4.x - 3 ) . ( 2x - 9 ) 

<=> -3.x² -x + 15x + 5 = -8.x² + 36x + 6x -27 

<=> 5x² -28x + 32 = 0

<=> \(\orbr{\begin{cases}x=4\\x=\frac{8}{5}\end{cases}}\)

Vậy x = \(\left\{4;\frac{8}{5}\right\}\)

Answer 6

4) Đặt : a = \(\sqrt{4-x}\left(a\ge0\right)\Leftrightarrow x=4-a^2\)

           :  b = \(\sqrt{x+1}\left(b\ge0\right)\Leftrightarrow b^2=x+1=5-a^2\)

<=> a² + b² = 5 

Phương trình trở thành : ( b² + 16 ) .a + ( 16 + a² ) .b -9ab = 36 

                      <=> a²b + ab² + 16. ( a + b ) - 9ab = 36 

                     <=> ( a + b ) .( ab +16 ) = 36 + 9ab = 9 .( 4 + ab ) 

                     <=> ( a² + b² + 2ab ) . ( a² b² + 32ab + 256 ) = 81 . ( 16 + 8ab +a² b² ) 

                            \---------/

                                 5 

Thay vào ta được : 5a² b² + 672ab + 2a³ b³ + 64a²b² + 1280=1296 + 648ab + 81a²b² 

                  <=> 2a³b³ - 12.a²b² + 24ab - 16  = 0 <=> ab = 2               (1) 

Từ ( 1 ) => ta có được : \(\sqrt{4-x}.\sqrt{x+1}=2\)

                                 <=> ( 4 - x ). ( x + 1 )                 = 4

                                  <=> -x² + 3x + 4                     =  4

                                 <=> -x²     + 3x                        = 0 

                                <=>\(\orbr{\begin{cases}x=0\\x=3\end{cases}\left(T/m\right)}\)

Vậy nghiệm cuối cùng là :{ 0 ; 3 } 

Answer 7

Trl :

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Answer 8

Trl:

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Answer 9

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Answer 10

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Answer 11

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