Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\sqrt{2x^2-16x+41}+\sqrt{3x^2-24x+64}=7\)
Ta đánh giá vế phải \(\sqrt{2x^2-16x+41}+\sqrt{3x^2-24x+64}=\sqrt{2\left(x-4\right)^2+9}+\sqrt{3\left(x-4\right)^2+16}\ge\sqrt{9}+\sqrt{16}=3+4=7\)(Do \(\left(x-4\right)^2\ge0\forall x\))
Như vậy, để \(\sqrt{2x^2-16x+41}+\sqrt{3x^2-24x+64}=7\)(hay dấu "=" xảy ra) thì \(\left(x-4\right)^2=0\)hay x = 4
Vậy nghiệm duy nhất của phương trình là 4
f, \(\sqrt{8+\sqrt{x}}+\sqrt{5-\sqrt{x}}=5\left(đk:25\ge x\ge0\right)\)
\(< =>\sqrt{8+\sqrt{x}}-\sqrt{9}+\sqrt{5-\sqrt{x}}-\sqrt{4}=0\)
\(< =>\frac{8+\sqrt{x}-9}{\sqrt{8+\sqrt{x}}+\sqrt{9}}+\frac{5-\sqrt{x}-4}{\sqrt{5-\sqrt{x}}+\sqrt{4}}=0\)
\(< =>\frac{\sqrt{x}-1}{\sqrt{8+\sqrt{x}}+\sqrt{9}}-\frac{\sqrt{x}-1}{\sqrt{5-\sqrt{x}}+\sqrt{4}}=0\)
\(< =>\left(\sqrt{x}-1\right)\left(\frac{1}{\sqrt{8+\sqrt{x}}+\sqrt{9}}-\frac{1}{\sqrt{5-\sqrt{x}}+\sqrt{4}}\right)=0\)
\(< =>x=1\)( dùng đk đánh giá cái ngoặc to nhé vì nó vô nghiệm )
a/ \(x^2+4x+5=2\sqrt{2x+3}\)
ĐK: \(x\ge-\frac{3}{2}\)
Cách 1:
Đặt \(\sqrt{2x+3}=y+2\text{ (}y\ge-2\text{)}\Rightarrow\left(y+2\right)^2=2x+3\text{ (1)}\)
Pt đã cho trở thành \(\left(x+2\right)^2+1=2\left(y+2\right)\Leftrightarrow\left(x+2\right)^2=2y+3\text{ (2)}\)
\(\left(2\right)-\left(1\right)\Rightarrow\left(x+2\right)^2-\left(y+2\right)^2=2\left(y-x\right)\Leftrightarrow\left(x-y\right)\left(x+y+6\right)=0\)
\(\Leftrightarrow x=y\text{ }\left(\text{do }x\ge-\frac{3}{2};\text{ }y\ge-2\text{ nên }x+y+6\ge-\frac{3}{2}-2+6>0\right)\)
Do đó, phương trình đã cho tương tương:
\(x=\sqrt{2x+3}-2\Leftrightarrow x+2=\sqrt{2x+3}\Leftrightarrow\left(x+2\right)^2=2x+3\)
\(\Leftrightarrow x^2+2x+1=0\Leftrightarrow\left(x+1\right)^2=0\Leftrightarrow x=-1\)
Kết luận: \(x=-1.\)
Cách 2:
\(pt\Leftrightarrow\frac{1}{4}\left(2x+3\right)^2+\frac{1}{2}\left(2x+3\right)+\frac{5}{4}=2\sqrt{2x+3}\)
Đặt \(t=\sqrt{2x+3};\text{ }t\ge0\)
pt thành \(\frac{1}{4}t^4+\frac{1}{2}t^3+\frac{5}{4}=2t\Leftrightarrow\left(t-1\right)^2\left(t^2+2t+5\right)=0\)
\(\Leftrightarrow t-1=0\text{ }\left(\text{do }t^2+2t+5=\left(t+1\right)^2+4>0\right)\)
\(\Leftrightarrow t=1\)
Do đó, phương trình đã cho tương đương:
\(\sqrt{2x+3}=1\Leftrightarrow x=-1\)
Kết luận: \(x=-1.\)
Cách 3:
\(pt\Leftrightarrow\left(x^2+2x+1\right)+\left[\left(2x+3\right)-2\sqrt{2x+3}+1\right]=0\)
\(\Leftrightarrow\left(x+1\right)^2+\left(\sqrt{2x+3}-1\right)^2=0\)
\(\Leftrightarrow x+1=0\text{ và }\sqrt{2x+3}-1=0\)
\(\Leftrightarrow x=-1\)
Kết luận: \(x=-1.\)
b/ \(2\left(x^2-3x+2\right)=3\sqrt{x^3+8}\)
ĐK: \(x\ge-2\)
\(pt\Leftrightarrow2\left(x^2-2x+4\right)-2\left(x+2\right)=3\sqrt{x+2}.\sqrt{x^2-2x+4}\)
Đặt \(a=\sqrt{x^2-2x+4};\text{ }b=\sqrt{x+2}\left(a>0;\text{ }b\ge0\right)\)
Pt thành: \(2a^2-2b^2=3ab\Leftrightarrow\left(a-2b\right)\left(2a+b\right)=0\)
\(\Leftrightarrow a=2b\text{ }\left(\text{do }a>0;\text{ }b\ge0\text{ nên }2a+b>0\right)\)
Pt đã cho tương đương: \(\sqrt{x^2-2x+4}=2\sqrt{x+2}\Leftrightarrow x^2-2x+4=4\left(x+2\right)\)
\(\Leftrightarrow x^2-6x-4=0\Leftrightarrow x=3+\sqrt{13}\text{ hoặc }x=3-\sqrt{13}\)
Kết luận: \(x=3+\sqrt{13};\text{ }x=3-\sqrt{13}\)
a) ĐK: \(x\ge5\)
\(\sqrt{4x-20}+\frac{1}{3}\sqrt{9x-45}-\frac{1}{5}\sqrt{16x-80}=0\)
\(\Leftrightarrow\)\(\sqrt{4\left(x-5\right)}+\frac{1}{3}\sqrt{9\left(x-5\right)}-\frac{1}{5}\sqrt{16\left(x-5\right)}=0\)
\(\Leftrightarrow\)\(2\sqrt{x-5}+\sqrt{x-5}-\frac{4}{5}\sqrt{x-5}=0\)
\(\Leftrightarrow\)\(\frac{11}{5}\sqrt{x-5}=0\)
\(\Leftrightarrow\)\(x-5=0\)
\(\Leftrightarrow\)\(x=5\) (t/m)
Vậy
b) \(-5x+7\sqrt{x}=-12\)
\(\Leftrightarrow\)\(5x-7\sqrt{x}-12=0\)
\(\Leftrightarrow\)\(\left(\sqrt{x}+1\right)\left(5\sqrt{x}-12\right)=0\)
đến đây tự làm
c) d) e) bạn bình phương lên
f) \(VT=\sqrt{3\left(x^2+2x+1\right)+9}+\sqrt{5\left(x^4-2x^2+1\right)+25}\)
\(=\sqrt{3\left(x+1\right)^2+9}+\sqrt{5\left(x^2-1\right)^2}\)
\(\ge\sqrt{9}+\sqrt{25}=8\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}x+1=0\\x^2-1=0\end{cases}}\)\(\Leftrightarrow\)\(x=-1\)
Vậy...
a) \(2\sqrt{3x}-4\sqrt{3x}+27-2\sqrt{3x}=27-4\sqrt{3x}\)
b) \(3\sqrt{2x}-5\sqrt{8x}+7\sqrt{8x}+28=3\sqrt{2x}+2\sqrt{8x}+28=3\sqrt{2x}+4\sqrt{2x}+28=7\sqrt{2x}+28\)
c) \(\frac{2}{x^2-y^2}\sqrt{\frac{3\left(x+y\right)^2}{2}}=\frac{2}{\left(x-y\right)\left(x+y\right)}.\frac{\sqrt{3}\left|x+y\right|}{\sqrt{2}}=\frac{\sqrt{6}}{x-y}\)
d) \(\frac{2}{2a-1}\sqrt{5a^2\left(1-4x+4a^2\right)}=\frac{2}{2a-1}\sqrt{5a^2\left(2a-1\right)^2}=\frac{2}{2a-1}.\sqrt{5}\left|a\left(2a-1\right)\right|=2a\sqrt{5}\)
Thiếu ĐKXĐ : ..............
a) Ta có: \(2\sqrt{3x}-4\sqrt{3x}+27-2\sqrt{3x}\)
\(=27-4\sqrt{3x}\)
b) Ta có: \(3\sqrt{2x}-5\sqrt{8x}+7\sqrt{8x}+28\)
\(=3\sqrt{2x}-5.2\sqrt{2x}+7.2\sqrt{2x}+28\)
\(=3\sqrt{2x}-10\sqrt{2x}+14\sqrt{2x}+28\)
\(=7\sqrt{2x}+28\)
c) Ta có: \(\frac{2}{x^2-y^2}.\sqrt{\frac{3\left(x+y\right)^2}{2}}\)
\(=\sqrt{\frac{4}{\left(x-y\right)^2.\left(x+y\right)^2}.\frac{3\left(x+y\right)^2}{2}}\)
\(=\sqrt{\frac{2.3}{\left(x-y\right)^2}}\)
\(=\frac{1}{x-y}.\sqrt{6}\)
d) Ta có: \(\frac{2}{2a-1}.\sqrt{5a^2.\left(1-4a+4a^2\right)}\)
\(=\sqrt{\frac{4}{\left(2a-1\right)^2}.5a^2.\left(2a-1\right)^2}\)
\(=2a.\sqrt{5}\)
Bài 2:
a, Ta có
\(3\sqrt{\left(-2\right)^2}+\sqrt{\left(-5\right)^2}\)
= \(3\left|-2\right|+\left|-5\right|\)
=\(6+5\)
= 11
Vậy \(3\sqrt{\left(-2\right)^2}+\sqrt{\left(-5\right)^2}=11\)
b, Ta có
\(\sqrt{6+2\sqrt{5}}-\sqrt{5}\)
= \(\sqrt{5+2\sqrt{5}+1}-\sqrt{5}\)
= \(\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{5}\)
= \(\left|\sqrt{5}+1\right|-\sqrt{5}\)
= \(\sqrt{5}+1-\sqrt{5}=1\)
Vậy \(\sqrt{6+2\sqrt{5}}-\sqrt{5}=1\)
a) \(\sqrt{x}+\sqrt{\frac{x}{9}}-\frac{1}{3}\sqrt{4x}=5\)
ĐK : x ≥ 0
<=>\(\sqrt{x}+\sqrt{x\times\frac{1}{9}}-\frac{1}{3}\sqrt{2^2x}=5\)
<=> \(\sqrt{x}+\sqrt{x\times\left(\frac{1}{3}\right)^2}-\left(\frac{1}{3}\times\left|2\right|\right)\sqrt{x}=5\)
<=> \(\sqrt{x}+\left|\frac{1}{3}\right|\sqrt{x}-\left(\frac{1}{3}\times2\right)\sqrt{x}=5\)
<=> \(\sqrt{x}+\frac{1}{3}\sqrt{x}-\frac{2}{3}\sqrt{x}=5\)
<=> \(\sqrt{x}\left(1+\frac{1}{3}-\frac{2}{3}\right)=5\)
<=> \(\sqrt{x}\times\frac{2}{3}=5\)
<=> \(\sqrt{x}=\frac{15}{2}\)
<=> \(x=\frac{225}{4}\)( tm )
a) \(\sqrt{3x-2}-\sqrt{2x+3}=\frac{3x-2-2x-3}{\sqrt{3x-2}+\sqrt{2x+3}}=\frac{x-5}{\sqrt{3x-2}+\sqrt{2x+3}}\)
\(\frac{x-5}{\sqrt{3x-2}+\sqrt{2x+3}}=\frac{x-5}{2}\Leftrightarrow\frac{x-5}{\sqrt{3x-2}+\sqrt{2x+3}}-\frac{x-5}{2}=0\)
\(\Leftrightarrow\left(x-5\right)\left(\frac{1}{\sqrt{3x-2}+\sqrt{2x+3}}-\frac{1}{2}\right)=0\). Do \(\frac{1}{\sqrt{3x-2}+\sqrt{2x+3}}-\frac{1}{2}\ne0\)
\(\Rightarrow x-5=0\Leftrightarrow x=5\). Vậy tập nghiệm của pt \(S=\left\{5\right\}\)
b) \(\sqrt{2}\left(x^2+8\right)=5\sqrt{x^3+8}\)
\(\Leftrightarrow x^2\sqrt{2}+8\sqrt{2}=5\sqrt{\left(x+2\right)\left(x^2-2x+4\right)}\)
Chắc cũng dùng trục căn thức ở mẫu nhưng mình chả biết làm thế nào :v
a, đk \(x\ge\frac{2}{3}\)
\(\sqrt{3x-2}-\sqrt{2x+3}=\frac{x-5}{2}\)
đặt \(\hept{\begin{cases}\sqrt{3x-2}=a\\\sqrt{2x+3}=b\end{cases}\left(a;b\ge0\right)}\)
pt trở thành : \(a-b=\frac{a^2-b^2}{2}\) \(\Leftrightarrow a^2-b^2=2a-2b\)
\(\Leftrightarrow a^2-2a-b^2+2b=0\)
\(\Leftrightarrow\left(a-1\right)^2-\left(b-1\right)^2=0\)
\(\Leftrightarrow\left(a-1-b+1\right)\left(a-1+b-1\right)=0\)
\(\Leftrightarrow\left(a-b\right)\left(a+b-2\right)=0\)
th1 : a - b = 0 <=> a = b hay \(\sqrt{3x-2}=\sqrt{2x+3}\)
\(\Leftrightarrow3x-2=2x+3\Leftrightarrow x=5\left(tm\right)\)
th2 : a + b - 2 = 0 hay \(\sqrt{3x-2}+\sqrt{2x+3}-2=0\)
\(\Leftrightarrow\sqrt{3x-2}=2-\sqrt{2x+3}\left(đk:x\le\frac{1}{2}\left(voli\right)\right)\)
vậy x = 5