chứng minh đẳng thức :
với a,b trái dấu \(\dfrac{b-a}{b\cdot\sqrt{\dfrac{-a}{b}}}=\dfrac{a-b}{a\cdot\sqrt{\dfrac{-b}{a}}}\)
Những câu hỏi liên quan
Cho hai biểu thức $A=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}$ và $B=\dfrac{1}{\sqrt{x}-2}+\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{2 \sqrt{x}}{4-x}$ vói $x \geq 0, x \neq 4$. a) Tính giá trị biểu thức $A$ với $x=1$. b) Chứng minh $B=\dfrac{\sqrt{x}}{\sqrt{x}+2}$ c) Tìm $x$ để $A \cdot B \geq 0$
1) Thay x=1x=1 vào biểu thức: A=√1+2√1−2A=1+21−2
A=−3A=−3
2) Chứng minh B=√x√x+2B=xx+2 với x≥0,x≠4x≥0,x≠4.
B=√x+2(√x−2)(√x+2)+(√x+1)(√x−2)(√x+2)(√x−2)−2√x(√x+2)(√x−2)B=x+2(x−2)(x+2)+(x+1)(x−2)(x+2)(x−2)−2x(x+2)(x−2)
=√x+2+x−√x−2−2√x(√x+2)(√x−2)=x−2√x(√x+2)(√x−2)=x+2+x−x−2−2x(x+2)(x−2)=x−2x(x+2)(x−2)
=√x(√x−2)(√x+2)(√x−2)=√x√x+2=x(x−2)(x+2)(x−2)=xx+2
3) Tìm xx để A⋅B≥0A⋅B≥0
A⋅B=√x+2√x−2⋅√x√x+2=√x√x−2A⋅B=x+2x−2⋅xx+2=xx−2.
TH1: x=0⇒√x=0⇒A⋅B=0x=0⇒x=0⇒A⋅B=0 (TM)
TH2: x>0⇒√x>0⇒√x−2>0⇒x>4x>0⇒x>0⇒x−2>0⇒x>4
Kết hợp điêu kiện: x=0x=0 hoặc x>4x>4 thỏa mãn yêu cầu.
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Thực hiện phép tính:
a) \(\left(\sqrt{ab}+2\sqrt{\dfrac{b}{a}}-\sqrt{\dfrac{a}{b}+\sqrt{\dfrac{1}{ab}}}\right)\cdot\sqrt{ab}\)
b) \(\left(\dfrac{am}{b}\sqrt{\dfrac{n}{m}}-\dfrac{ab}{n}\sqrt{mn}+\dfrac{a^2}{b^2}\sqrt{\dfrac{m}{n}}\right)\cdot a^2b^2\cdot\sqrt{\dfrac{n}{m}}\)
a: \(=ab+2\cdot\sqrt{\dfrac{b}{a}\cdot ab}-\sqrt{ab\cdot\left(\dfrac{a}{b}+\dfrac{1}{\sqrt{ab}}\right)}\)
\(=ab+2b-\sqrt{ab\cdot\dfrac{a\sqrt{a}+\sqrt{b}}{b\sqrt{a}}}\)
\(=ab+2b-\sqrt{\sqrt{a}\cdot\left(a\sqrt{a}+\sqrt{b}\right)}\)
b: \(=\left(\sqrt{\dfrac{a^2m^2\cdot n}{b^2\cdot m}}-\sqrt{mn\cdot\dfrac{a^2b^2}{n^2}}+\sqrt{\dfrac{a^4}{b^4}\cdot\dfrac{m}{n}}\right)\cdot a^2b^2\cdot\sqrt{\dfrac{n}{m}}\)
\(=\left(\dfrac{a\sqrt{mn}}{b}-\sqrt{a^2b^2\cdot\dfrac{m}{n}}+\dfrac{a^2}{b^2}\cdot\sqrt{\dfrac{m}{n}}\right)\cdot\sqrt{\dfrac{n}{m}}\cdot a^2b^2\)
\(=\left(\dfrac{an}{b}-ab+\dfrac{a^2}{b^2}\right)\cdot a^2b^2\)
\(=a^3nb-a^3b^3+a^4\)
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Chứng minh các đẳng thức sau:
c) \(\dfrac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{a}+2\sqrt{b}}-\dfrac{2b}{b-a}=\dfrac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}\) ( với a,b > 0 và a \(\ne\) b )
\(\dfrac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{a}+2\sqrt{b}}-\dfrac{2b}{b-a}\left(a,b>0;a\ne b\right)\\ =\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2-\left(\sqrt{a}-\sqrt{b}\right)^2+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\\ =\dfrac{4\sqrt{ab}+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\\ =\dfrac{4\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}=\dfrac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)
Tick plz
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Ta có: \(\dfrac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{a}+2\sqrt{b}}-\dfrac{2b}{b-a}\)
\(=\dfrac{a+2\sqrt{ab}+b-a+2\sqrt{ab}-b+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{4b+4\sqrt{ab}}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{4\sqrt{b}\left(\sqrt{b}+\sqrt{a}\right)}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{b}+\sqrt{a}\right)}\)
\(=\dfrac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)
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Bài 75 (trang 40 SGK Toán 9 Tập 1)
Chứng minh các đẳng thức sau:
a) $left(dfrac{2 sqrt{3}-sqrt{6}}{sqrt{8}-2}-dfrac{sqrt{216}}{3}right) cdot dfrac{1}{sqrt{6}}-1,5$;
b) $left(dfrac{sqrt{14}-sqrt{7}}{1-sqrt{2}}+dfrac{sqrt{15}-sqrt{5}}{1-sqrt{3}}right): dfrac{1}{sqrt{7}-sqrt{5}}-2$;
c) $dfrac{a sqrt{b}+b sqrt{a}}{sqrt{a b}}: dfrac{1}{sqrt{a}-sqrt{b}}a-b$ với $a, b$ dương và $a neq b$;
d) $left(1+dfrac{a+sqrt{a}}{sqrt{a}+1}right)left(1-dfrac{a-sqrt{a}}{sqrt{a}-1}right)1-a$ với $a geq 0$ và $a ne...
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Bài 75 (trang 40 SGK Toán 9 Tập 1)
Chứng minh các đẳng thức sau:
a) $\left(\dfrac{2 \sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right) \cdot \dfrac{1}{\sqrt{6}}=-1,5$;
b) $\left(\dfrac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\dfrac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right): \dfrac{1}{\sqrt{7}-\sqrt{5}}=-2$;
c) $\dfrac{a \sqrt{b}+b \sqrt{a}}{\sqrt{a b}}: \dfrac{1}{\sqrt{a}-\sqrt{b}}=a-b$ với $a, b$ dương và $a \neq b$;
d) $\left(1+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\right)=1-a$ với $a \geq 0$ và $a \neq 1$.
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a) \(\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right).\dfrac{1}{\sqrt{6}}=\left(\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-\dfrac{\sqrt{216}}{3}\right).\dfrac{1}{\sqrt{6}}=\dfrac{1}{2}-2=-\dfrac{3}{2}\)
Xem thêm câu trả lời
chứng minh răng :
a,\(\left(1+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\right)\cdot\left(1-\dfrac{a-\sqrt{a}}{a-1}\right)=1-a\left(a>hoaăặc=0,a\right)\left(a#1\right)\)b, \(\dfrac{\sqrt{ab}-b}{\sqrt{b}}-\sqrt{\dfrac{a}{b}}< 0\left(a>hoac=0,b>0\right)\)
\(=\dfrac{\sqrt{ab}-b-\sqrt{a}}{\sqrt{b}}\)
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1. chứng minh rằng các hằng đẳng thức sau với điều kiện các biểu thức tồn tại:
a) \(\dfrac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}:\dfrac{1}{\sqrt{a}-\sqrt{b}}=a-b\)
b)\(\left(1+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\right)=1-a\)
a, \(VT=\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}.\left(\sqrt{a}-\sqrt{b}\right)=a-b=VP\) đpcm
b,\(VT=1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}-\dfrac{a^2-a}{a-1}=1-\sqrt{a}+\sqrt{a}-a=1-a=VP\) đpcm
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Chứng minh các đẳng thức sau:
a) \(\left(1+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\right)\left(1-\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\right)=1-x\)
(Với \(x\ge0;x\ne1\))
b) \(\dfrac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}+\dfrac{a-b}{\sqrt{a}-b}=2\sqrt{a}\)
(Với a>0; b>0; \(a\ne b\))
Câu b bạn sửa lại đề
\(a,VT=\left[1+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right]\left[1-\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right]\\ =\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)=1-x=VP\\ b,VT=\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}+\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\\ =\sqrt{a}-\sqrt{b}+\sqrt{a}+\sqrt{b}=2\sqrt{a}=VP\)
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a: \(=\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)=1-x\)
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cho tỉ lệ thức \(\dfrac{a}{b}=\dfrac{c}{d}\)
chứng minh
a. \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
b. \(\dfrac{a\cdot b}{c\cdot d}=\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
c.\(\dfrac{2008\cdot a-2009\cdot b}{2009\cdot c+2010\cdot d}=\dfrac{2008\cdot c-2009\cdot d}{2009\cdot a+2010\cdot b}\)
a: Đặt a/b=c/d=k
=>a=bk; c=dk
\(\left(\dfrac{a+b}{c+d}\right)^2=\left(\dfrac{bk+b}{dk+d}\right)^2=\dfrac{b^2}{d^2}\)
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{b^2k^2+b^2}{d^2k^2+d^2}=\dfrac{b^2}{d^2}\)
Do đó: \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
b: \(\dfrac{ab}{cd}=\dfrac{bk\cdot b}{dk\cdot d}=\dfrac{b^2}{d^2}\)
\(\left(\dfrac{a-b}{c-d}\right)^2=\left(\dfrac{bk-b}{dk-d}\right)^2=\dfrac{b^2}{d^2}\)
Do đó: \(\dfrac{ab}{cd}=\left(\dfrac{a-b}{c-d}\right)^2\)
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cho tỉ lệ thức \(\dfrac{a}{b}=\dfrac{c}{d}\)
chứng minh \(\dfrac{4\cdot a-5\cdot b}{4\cdot a+5\cdot b}=\dfrac{4\cdot c-5\cdot d}{4\cdot c+5\cdot d}\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=b.k\\c=d.k\end{matrix}\right.\)
Ta có:
\(\dfrac{4.a-5.b}{4.a+5.b}=\dfrac{4.a+5.b-10.b}{4.a+5.b}=1-\dfrac{10.b}{4.a+5.b}=1-\dfrac{10.b}{4.b.k+5b}=1-\dfrac{10}{4.k+5}\) (1)
\(\dfrac{4.c-5.d}{4.c+5.d}=\dfrac{4.c+5.d-10.d}{4.c+5.d}=1-\dfrac{10.d}{4.c+5.d}=1-\dfrac{10.d}{4.d.k+5.d}=1-\dfrac{10}{4.k+5}\) (2)
Từ (1) và (2) suy ra \(\dfrac{4.a-5.b}{4.a+5.b}=\dfrac{4.c-5.d}{4.c+5.d}\left(đpcm\right)\)
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Lời giải:
Đặt \(\frac{a}{b}=\frac{c}{d}=t\Rightarrow a=bt; c=dt\)
Khi đó ta có:
\(\frac{4a-5b}{4a+5b}=\frac{4bt-5b}{4bt+5b}=\frac{b(4t-5)}{b(4t+5)}=\frac{4t-5}{4t+5}\)
\(\frac{4c-5d}{4c+5d}=\frac{4dt-5d}{4dt+5d}=\frac{d(4t-5)}{d(4t+5)}=\frac{4t-5}{4t+5}\)
Do đó: \(\frac{4a-5b}{4a+5b}=\frac{4c-5d}{4c+5d}\) (đpcm)
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